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Calculus 2 : Integrals

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Example Questions

Example Question #21 : Integrals

Find the indefinite integral.

$\int x^2\sin(x)dx$

Possible Answers:

$-x^2\cos(x)+2x\cos(x)+C$

$-\frac{1}{3}x^3\cos(x)+C$

$-x^2\sin(x)+2x\sin(x)+2\sin(x)+2+C$

$-x^2\cos(x)+2x\sin(x)+2\cos(x)+C$

Correct answer:

$-x^2\cos(x)+2x\sin(x)+2\cos(x)+C$

Explanation:

Use Integration By Parts (IBP) to simplify the integral, using the IBP formula

$\int udv = uv - \int vdu$

Assign the variables and to appropriate expressions from the original integral,

$u = x^{2} \rightarrow du = 2xdx$

$dv = \sin(x)dx \rightarrow v = -\cos(x)$

Substitute these new expressions into the IBP formula and our integral becomes as follows,

$\int x^2\sin(x)dx = -x^{2}\cos(x)- \int -\cos(x)2xdx$

$= -x^{2}\cos(x) + \int \cos(x)2xdx$

Use IBP again on the integral $\int \cos(x)2xdx$ and assign the variables and to appropriate expressions from the integral,

$u = 2x \rightarrow du= 2dx$

$dv = \cos(x)dx \rightarrow v= \sin(x)$

Substitute these new expressions into the IBP formula again and our full expression becomes as follows,

$-x^{2}\cos(x) + \left [ 2x\sin(x)-\int 2\sin(x)dx \right ]$

Integrate using the trig integration rules as follows,

$\int \sin(x)dx = - \cos(x)$

So the final indefinite integral expression becomes

$-x^{2}\cos(x) + 2x\sin(x)- (-2\cos(x)) +C$

And the final answer is as follows,

$-x^{2}\cos(x) + 2x\sin(x)+ 2\cos(x) +C$

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Example Question #21 : Integrals

Find the indefinite integral.

$\int \frac{1}{x^2+x\sqrt{x}}dx$

Possible Answers:

$ln\left | x^2+x\sqrt{x}\right | + C$

$-2ln\left | \sqrt{x} \right |-\frac{2}{\sqrt{x}}+2ln\left | \sqrt{x}+1 \right | + C$

$2ln\left | x^\frac{3}{2}+x\right | + C$

$-2ln\left | x^\frac{3}{2}\right | + 2ln\left | x\right | + C$

$ln\left | \sqrt{x} \right |+\sqrt{x}+ln\left | \frac{3}{\sqrt{x}} \right | + C$

Correct answer:

$-2ln\left | \sqrt{x} \right |-\frac{2}{\sqrt{x}}+2ln\left | \sqrt{x}+1 \right | + C$

Explanation:

Make an appropriate u-substitution to simplify the integral.

$u = \sqrt{x} \rightarrow u^{2}=x \rightarrow u^{4}= x^{2}$

$du = \frac{1}{2}x^{-\frac{1}{2}}dx = \frac{1}{2\sqrt{x}}dx$

$2\sqrt{x}du=dx$

Since $u = \sqrt{x}$ ,

2udu=dx

Substitute the new expressions containing and back into the original integral as follows,

$\int \frac{1}{u^{4}+u^{3}}2udu \Leftrightarrow 2\int \frac{u}{u(u^{3}+u^{2})}du \Leftrightarrow 2\int \frac{1}{u^{3}+u^{2}}du$

Use partial fraction decomposition on the integral to simplify the expression as follows,

$\frac{1}{u^{3}+u^{2}} = \frac{1}{u^{2}(u+1)} = \frac{A}{u}+ \frac{B}{u^{2}}+ \frac{C}{u+1}$

$1=A(u)(u+1)+B(u+1)+C(u^{2})$

Set u = 0 then,

$1 = A(0)(0+1)+B(0+1)+C(0^{2}) \rightarrow B=1$

Set u = -1 then,

$1 = A(-1)(-1+1)+B(-1+1)+C(-1)^{2} \rightarrow C=1$

Set u = 1 then,

$1 = A(1)(1+1)+B(1+1)+C(1)^{2}$

1 = A(2)+ (1)(2)+ (1)(1)

$1 = 2A + 2 + 1 \rightarrow A = -1$

So,

$\frac{1}{u^{3}+u^{2}} = \frac{-1}{u}+ \frac{1}{u^{2}}+ \frac{1}{u+1}$

And our integral expression becomes

$2\int (\frac{-1}{u}+ \frac{1}{u^{2}}+ \frac{1}{u+1})du$

Integrate each term. We will need to use the power rule and natural log integration as follows,

$\int \frac{1}{x}dx = ln\left | x\right | \rightarrow \int \frac{1}{u+1}du = ln\left | u + 1\right |$

Where $n\neq 1$ , $\int \frac{1}{x^{n}}dx = \int x^{-n}dx = \frac{x^{-n+1}}{-n+1} \rightarrow \int \frac{1}{u^{2}}du = -\frac{1}{u}$

This results in our final indefinite integral,

$2(-ln\left | u\right |-\frac{1}{u}+ln\left | u + 1\right |) + C$

Substitute back in our original u-substitution $u = \sqrt{x}$ and simplify to give us our final answer

$-2ln\left | \sqrt{x}\right |-\frac{2}{\sqrt{x}}+2ln\left | \sqrt{x}+ 1\right | + C$

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Example Question #23 : Integrals

Find the definite integral.

$\int_{0}^{8\pi ^2}\cos(\sqrt{x+\pi ^2})dx$

Possible Answers:

$-3\pi$

Correct answer:

Explanation:

Simplify the integral using an appropriate u-substitution,

$u = \sqrt{x+\pi^{2}}$

$du = \frac{1}{2}(x+\pi^{2})^{-\frac{1}{2}}dx = \frac{1}{2\sqrt{x+\pi^{2}}}dx$

$2\sqrt{x+\pi^{2}}du=dx$

Since $u = \sqrt{x+\pi^{2}}$ ,

2udu=dx

Substitute the new expressions containing and into the original integral as follows,

$\int \cos(u)2udu$

$2\int u \cos(u)du$

Use Integration By Parts (IBP) to simplify the integral further, using the IBP formula

$\int wdv = wv - \int vdw$

Assign the variables and to appropriate expressions from the integral

$w = u \rightarrow dw = du$

$dv = \cos(u)du \rightarrow v = \sin(u)$

Substitute these expressions back into the IBP formula and our integral becomes as follows,

$2\left ( u\sin(u)-\int \sin(u)du\right )$

Integrate using the trig integration rule

$\int \sin(x)dx = -\cos(x)$

Our indefinite integral becomes

$2(u\sin(u)-(-\cos(u))) +C \Leftrightarrow 2u\sin(u)+2\cos(u)+C$

Substitute back in our original u-substitution $u = \sqrt{x+\pi^{2}}$ to get our final indefinite integral,

$2\sqrt{x+\pi^{2}}\sin(\sqrt{x+\pi^{2}})+2\cos(\sqrt{x+\pi^{2}})+C$

Apply the given boundaries x = 0 to $x = 8\pi^2$ to get our final answer.

$\left [ 2\sqrt{8\pi^2+\pi^{2}}\sin(\sqrt{8\pi^2+\pi^{2}})+2\cos(\sqrt{8\pi^2+\pi^{2}})\right ]-\left [ 2\sqrt{0+\pi^{2}}\sin(\sqrt{0+\pi^{2}})+2\cos(\sqrt{0+\pi^{2}})\right ]$

$\left [ 2(3\pi)(0)+2(-1)\right ] - \left [ 2(\pi)(0)+2(-1)\right ]$

$\left [ -2\right ]- \left [ -2\right ] = 0$

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Example Question #21 : Integrals

Find the indefinite integral.

$\int x(x+7)^{100} dx$

Possible Answers:

$\frac{1}{2}x^2+\frac{1}{101}(x+7)^{101}+C$

$\frac{1}{102}x^{102}-\frac{7}{102}x^{101}+C$

$\frac{(x^2+7x)^{101}}{101}*(\frac{1}{3}x^3+\frac{7}{2}x^2)+C$

$\frac{1}{102}(x+7)^{102}-\frac{7}{101}(x+7)^{101}+C$

Correct answer:

$\frac{1}{102}(x+7)^{102}-\frac{7}{101}(x+7)^{101}+C$

Explanation:

Make an appropriate u-substitution to transform the integral into something we are familiar with.

$u = x+7 \rightarrow u-7=x$

du = dx

Substitute the new expressions containing and into the original equation as follows,

$\int (u-7)(u)^{100}du$

$\int \left ( u^{101}-7u^{100}\right )du$

Integrate each term using the power rule.

$\int cx^{n}dx = \frac{cx^{n+1}}{n+1} \rightarrow \int -7u^{100}du = -\frac{7u^{101}}{101}$

After integration we get the expression as follows,

$\frac{u^{102}}{102}-\frac{7u^{101}}{101}+C$

Substitute back in our original u-substitution u = x+7 to get the final answer.

$\frac{(x+7)^{102}}{102}-\frac{7(x+7)^{101}}{101}+C$

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Example Question #25 : Integrals

Find the indefinite integral.

$\int x^2ln\left | x+8\right |dx$

Possible Answers:

$\frac{1}{3}x^3ln\left | x+8\right |-\frac{1}{12}x^4ln\left | x+8\right |+C$

$\frac{1}{3}x^3ln\left | x+8\right |-\frac{1}{9}x^3+\frac{4}{3}x^2-\frac{64}{3}x+\frac{512}{3}ln\left | x+8\right |+C$

$\frac{1}{3}x^3ln\left | x+8\right |+C$

$\frac{1}{3(x+8)}x^3+C$

Correct answer:

$\frac{1}{3}x^3ln\left | x+8\right |-\frac{1}{9}x^3+\frac{4}{3}x^2-\frac{64}{3}x+\frac{512}{3}ln\left | x+8\right |+C$

Explanation:

First use Integration By Parts (IBP) to make the integral doable, using the IBP formula,

$\int udv = uv-\int vdu$

Assign appropriate expressions from the original integral to the variables and to use in the IBP formula.

$u = ln\left | x+8\right |\rightarrow du = \frac{1}{x+8}dx$

$dv = x^{2}dx \rightarrow v = \frac{1}{3}x^{3}$

Substitute these expressions into the IBP formula and our integral becomes as follows,

$\int x^2ln\left | x+8\right |dx = \frac{1}{3}x^{3}ln\left | x+8\right |- \int \frac{1}{3}x^3*\frac{1}{x+8}dx$

Use long division to simplify the expression $\frac{x^{3}}{x+8}$ ,

$x^{2} -8x+64-\frac{512}{x+8}$

$x+8|\overline{x^3 \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; }$

$\\ -(x^{3}+8x^{2})\\ \overline{\ \ \ \ \ \ \ \ -8x^{2}\ \ \ \ }\\ {\ \ \ \ \ -(-8x^{2}-64x)}\\ \overline{\ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ 64x}\\ {\ \ \ \ \ \ \ \ \ \ \ \ \ \ \ -(64x+512)}\\ \overline{\ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ 512}$

So,

$\frac{x^{3}}{x+8} = x^{2} -8x+64-\frac{512}{x+8}$

Substitute this back into its appropriate place in our expression and it becomes as follows,

$\frac{1}{3}x^{3}ln\left | x+8\right |- \frac{1}{3} \int (x^{2} -8x+64-\frac{512}{x+8})dx$

Integrate each term using the power rule and natural log integration as follows,

$\int x^{n}dx = \frac{x^{n+1}}{n+1} \rightarrow \int x^{2}dx= \frac{1}{3}x^{3}$

$\int \frac{c}{x}dx = cln\left | x\right | \rightarrow \int \frac{-512}{x+8}dx = -512ln\left | x+8\right |$

Our indefinite integral becomes as follows,

$\frac{1}{3}x^{3}ln\left | x+8\right |- \frac{1}{3}(\frac{1}{3}x^{3}-4x^{2}+64x-512ln\left | x+8\right |)+C$

Simplifying, our answer is

$\frac{1}{3}x^{3}ln\left | x+8\right |- \frac{1}{9}x^{3}-\frac{4}{3}x^{2}+\frac{64}{3}x-\frac{512}{3}ln\left | x+8\right |+C$

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Example Question #21 : Introduction To Integrals

Find the indefinite integral.

$\int \frac{x^3-4x+7}{x^2+2x-3}dx$

Possible Answers:

$\frac{\frac{1}{4}x^4-2x^2+7x}{\frac{1}{3}x^3+x^2-3x}+C$

$\frac{1}{2}x^2-2x+ln\left | x-1\right |+2ln\left | x+3\right | + C$

$2ln\left | x-1\right |+ln\left | x+3\right | + C$

$\frac{1}{2}x^2-2x-\frac{7}{3}x + C$

Correct answer:

$\frac{1}{2}x^2-2x+ln\left | x-1\right |+2ln\left | x+3\right | + C$

Explanation:

First divide the numerator by the denominator, using long division.

$x-2 +\frac{3x+1}{x^2+2x-3}{}$

$x^2+2x-3|\overline{x^3-4x+7}$

$\\ -x^3-2x^2-3x\\ \overline{-2x^2-x+7\ \ \ \ }\\+2x^2+4x-6\\ \overline{\ \ \ \ \ \ \ \ \ \ \ 3x-1}$

Now we can write our integral as follows.

$\int x-2 +\frac{3x+1}{(x+3)(x-1)}dx$

From here we will use partial fraction decomposition.

$\frac{3x+1}{(x+3)(x-1)}=\frac{A}{(x+3)}+\frac{B}{x-1}$

3x+1=A(x-1)+B(x+3)

Set x=1 then,

$4=A(0)+b(4)\rightarrow B=1$ .

Set x=-3 then,

$-8=A(-4)+B(0)\rightarrow A=2$

Therefore we get,

$\frac{3x+1}{(x+3)(x-1)}=\frac{2}{(x+3)}+\frac{1}{(x-1)}$ .

Now we can rewrite our integral as follows,

$\int x-2 +\frac{2}{x+2}+\frac{1}{x-1}dx$ .

From here integrate each term. To integrate we will need to use the power rule, integration of a constant, and nautral log integration we are as follows.

$\\ \int x dx =\frac{1}{2}x^2 \rightarrow \int x^n=\frac{1}{n+1}x^{n+1}\\ \int 2dx=2x \rightarrow \int c\ dx=cx \\ \int \frac{2}{x+3}dx=2ln|x+3| \rightarrow \int\frac{dx}{x}=ln|x|$

Resulting in our final answer,

$\frac{1}{2}x^2-2x+ln\left | x-1\right |+2ln\left | x+3\right | + C$ .

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Example Question #27 : Integrals

Find the indefinite integral.

$\int \frac{x^4+7}{x^3+x^2}dx$

Possible Answers:

$\frac{1}{2}x^2-x-7ln\left | x\right |-\frac{7}{x}+8ln\left | x+1\right |+C$

$ln\left | x\right |-\frac{28}{x^4}+C$

$(\frac{1}{5}x^5+7x)*ln\left | x^3+x^2\right |+C$

$\frac{1}{3}x^3ln\left | x+1\right |+7ln\left | x^3+x^2\right |+C$

Correct answer:

$\frac{1}{2}x^2-x-7ln\left | x\right |-\frac{7}{x}+8ln\left | x+1\right |+C$

Explanation:

Simplify the integrand using long division.

$x - 1 + \frac{x^2 +7}{x^3 + x^2}$

$x^3+x^2|\overline{x^4 + 7 \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \;}$

$\\ -(x^4+x^3)\\ \overline{\ \ \ \ \ \ \ \ \ -x^3\ \ \ \ }\\ {\ \ \ \ \ -(-x^3-x^2)}\\ \overline{\ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ x^2+7}$

The integral becomes

$\int \left ( x - 1 + \frac{x^2 +7}{x^3 + x^2} \right ) dx$

Use partial fraction decomposition to simplify the expression $\frac{x^2 +7}{x^3 + x^2}$ ,

$\frac{x^2 +7}{x^3 + x^2} = \frac{x^2 + 7}{x^2(x+1)} = \frac{A}{x}+ \frac{B}{x^2}+\frac{C}{x+1}$

x^2 + 7 = A(x)(x+1)+B(x+1)+C(x^2)

Set x = 0 then,

$(0)^2+7 = A(0)(0+1)+B(0+1)+C(0^2) \rightarrow B = 7$

Set x = -1 then,

$(-1)^2 +7 = A(-1)(-1+1)+B(-1+1)+C(-1)^2 \rightarrow C = 8$

Set x = 1 then,

(1)^2 + 7 = A(1)(1+1)+B(1+1)+C(1^2)

8 = A(2)+ (7)(2)+ (8)(1)

$8 = 2A + 14 + 8 \rightarrow A = -7$

Therefore we get

$\frac{x^2 +7}{x^3 + x^2} = \frac{-7}{x}+ \frac{7}{x^2}+\frac{8}{x+1}$

Now we can rewrite our integral as follows,

$\int \left ( x - 1 + \frac{-7}{x}+ \frac{7}{x^2}+\frac{8}{x+1}\right ) dx$

Integrate each term using the power rule, integration of a constant, and natural log integration as follows,

$\int x^n=\frac{1}{n+1}x^{n+1} \rightarrow \int xdx = \frac{x^2}{2}$

$\int c\ dx=cx \rightarrow \int 1dx = x$

$\int\frac{c}{x}dx=cln|x| \rightarrow \int -\frac{7}{x}dx = -7ln\left | x\right |$

Resulting in our final answer,

$\frac{1}{2}x^2 - x -7ln\left | x\right | - \frac{7}{x} + 8ln\left | x+1\right |+C$

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Example Question #28 : Integrals

Find the indefinite integral.

$\int \frac{ln\left | x\right |}{(x+7)^2}dx$

Possible Answers:

$-\frac{ln\left | x\right |}{x+7}+\frac{1}{7}ln\left | x\right |-\frac{1}{7}ln\left | x+7\right |+C$

$ln\left | x\right |+2ln\left | x+7\right |+C$

$\frac{1}{2x^2+14x}+C$

$xln\left | x\right |-\frac{1}{x+7}+C$

Correct answer:

$-\frac{ln\left | x\right |}{x+7}+\frac{1}{7}ln\left | x\right |-\frac{1}{7}ln\left | x+7\right |+C$

Explanation:

Use Integration By Parts (IBP) to make the integral doable, using the IBP formula.

$\int udv = uv - \int vdu$

Assign appropriate expressions from the original integral to the variables and to use in the IBP formula.

$u = ln\left | x\right | \rightarrow du = \frac{1}{x}dx$

$dv = (x+7)^{-2}dx \rightarrow v = -(x+7)^{-1}$

Substitute these expressions into the IBP formula and our integral becomes as follows,

$-(x+7)^{-1}ln\left | x\right | - \int -(x+7)^{-1}\frac{1}{x}dx$

$-\frac{1}{x+7}ln\left | x\right |+ \int \frac{1}{x(x+7)}dx$

Use partial fraction decomposition to further simplify

$\frac{1}{x(x+7)} = \frac{A}{x} + \frac{B}{x+7}$

1 = A(x+7)+B(x)

Set x = 0 then,

$1 = A(0+7)+B(0) \rightarrow 1 = 7A \rightarrow A = \frac{1}{7}$

Set x = -7 then,

$1 = A(-7+7)+B(-7) \rightarrow 1 = -7B \rightarrow B = -\frac{1}{7}$

So,

$\frac{1}{x(x+7)} = \frac{\frac{1}{7}}{x} + \frac{-\frac{1}{7}}{x+7}$

Therefore we get

$-\frac{1}{x+7}ln\left | x\right |+ \int \left ( \frac{\frac{1}{7}}{x} + \frac{-\frac{1}{7}}{x+7}\right ) dx$

Finish the integration using the natural log integration rule

$\int \frac{c}{x}dx = cln\left | x\right | \rightarrow \int \frac{\frac{1}{7}}{x}dx = \frac{1}{7}ln\left | x\right |$

Resulting in the final answer,

$-\frac{1}{x+7}ln\left | x\right |+ \frac{1}{7}ln\left | x\right | -\frac{1}{7}ln\left | x + 7\right |+ C$

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Example Question #21 : Integrals

Find the indefinite integral.

$\int x^2\sqrt{x^3+5} dx$

Possible Answers:

$\frac{3}{2}(x^5+5)^\frac{2}{3} + C$

$\frac{1}{3}x^3+ (x^3+5)^\frac{3}{2} + C$

$\frac{2}{9}(x^3+5)^\frac{3}{2} + C$

$\frac{2}{9}x^\frac{3}{2} + C$

Correct answer:

$\frac{2}{9}(x^3+5)^\frac{3}{2} + C$

Explanation:

Simplify the integral choosing an appropriate u-substitution

$u = x^{3}+5$

$du = 3x^{2}dx\rightarrow \frac{1}{3}du=x^{2}dx$

Substitute the new expressions containing back into the original integral as follows,

$\int \frac{1}{3}du\sqrt{u}$

Since $\sqrt{u} = u^{\frac{1}{2}}$ ,

$\frac{1}{3}\int \sqrt{u}du = \frac{1}{3}\int u^{\frac{1}{2}}du$

Integrate using the power rule of integration,

$\int x^{n}dx = \frac{x^{n+1}}{n+1} \rightarrow \int u^{\frac{1}{2}}du = \frac{2u^{\frac{3}{2}}}{3}$

Resulting in the answer after integrating,

$\frac{1}{3}\left [ \frac{2}{3}u^{\frac{3}{2}} \right ] +C$

$\frac{2}{9}u^{\frac{3}{2}}+C$

Substitute back in our original $u = x^{3}+5$ for our final answer,

$\frac{2}{9}\left ( x^{3}+5\right )^{\frac{3}{2}}+C$

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Example Question #21 : Integrals

Find the indefinite integral.

$\int e^{\sqrt{x}}dx$

Possible Answers:

$xe^{\sqrt{x}} + C$

$2e^{\sqrt{x}}(\sqrt{x}-1)+C$

$2e^{\sqrt{x}} + C$

$\bigca\sqrt{x}e^{\sqrt{x}}+ C$

Correct answer:

$2e^{\sqrt{x}}(\sqrt{x}-1)+C$

Explanation:

Make an appropriate u-substitution to turn the integral into something we are familiar with,

$u=\sqrt{x}$

$du = \frac{1}{2}x^{-\frac{1}{2}}dx =\frac{1}{2} *\frac{1}{\sqrt{x}}dx$

$2\sqrt{x}du = dx$ .

And since $u = \sqrt{x}$ ,

2udu =dx

Substitute the new expressions containing back into the original integral as follows,

$\int e^{u}2udu$

$2\int ue^{u}du$ .

From here we will use Integration By Parts (IBP) to simplify the integral further, using the IBP formula

$\int wdv = wv - \int vdw$ .

Assign the variables and to appropriate expressions from the original integral,

$w = u \rightarrow dw = du$

$dv = e^{u}du \rightarrow v = e^{u}$ .

Substitute these new expressions back into the IBP formula and our integral becomes as follows,

$2\int ue^{u}du = 2(ue^{u}-\int e^{u}du)$ .

Integrate using the following integration rule:

$\int e^{x}dx = e^{x} \rightarrow \int e^{u}du = e^{u}$ .

This results in the expression

$2(ue^{u}-e^{u}) +C$ .

Substitute back in our original u-substitution $u = \sqrt{x}$ ,

$2(\sqrt{x}e^{\sqrt{x}}-e^{\sqrt{x}}) +C$ .

Simplify to get the final answer,

$2e^{\sqrt{x}}(\sqrt{x}-1) +C$ .

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Calculus 2 : Integrals

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