\(\displaystyle \begin{align*}&\text{In performing an integral, familiarity with derivative rules}\\&\text{can prove useful. After all, an integral is essentially the}\\&\text{opposite operation of a derivative. Consider how in a derivative}\\&\text{we’d use the chain rule to multiply the derivative of the “inside”}\\&\text{of a function by the derivative of the outside? Integrals, in}\\&\text{similar fashion, entail a division.}\\&\text{Utilizing integral rules:}\\&\int[cos(ax)]=\frac{sin(ax)}{a}\\&\int[sin(ax)]=-\frac{cos(ax)}{a}\\&\text{Note in deriving the inside, the lone constant goes to zero.}\\&\text{Taking the first integral we find:}\\&\frac{(3sin(x + 5))}{79}+C_{1}\\&\text{Note that after indefinite integration we add a constant, }C_1\\&\text{Integrating again, we'll add a new constant, while multiplying}\\&\text{our first one by }x:\\&-\frac{(3cos(x + 5))}{79}+C_{1}x+C_{2}\\&\end{align*}\)

\(\displaystyle \begin{align*}&\text{In performing an integral, familiarity with derivative rules}\\&\text{can prove useful. After all, an integral is essentially the}\\&\text{opposite operation of a derivative. Consider how in a derivative}\\&\text{we’d use the chain rule to multiply the derivative of the “inside”}\\&\text{of a function by the derivative of the outside? Integrals, in}\\&\text{similar fashion, entail a division.}\\&\text{Utilizing integral rules:}\\&\int x^{a}=\frac{x^{a+1}}{a+1}\\&\text{Note for this problem, the }a\text{ value is } 16\\&\text{Taking the first integral we find:}\\&\frac{(7x^{17})}{51}+C_{1}\\&\text{Note that after indefinite integration we add a constant, }C_1\\&\text{Integrating again, we'll add a new constant, while multiplying}\\&\text{our first one by }x:\\&\frac{(7x^{18})}{918}+C_{1}x+C_{2}\\&\end{align*}\)

\(\displaystyle \begin{align*}&\text{In performing an integral, familiarity with derivative rules}\\&\text{can prove useful. After all, an integral is essentially the}\\&\text{opposite operation of a derivative. Consider how in a derivative}\\&\text{we’d use the chain rule to multiply the derivative of the “inside”}\\&\text{of a function by the derivative of the outside? Integrals, in}\\&\text{similar fashion, entail a division.}\\&\text{Utilizing integral rules:}\\&\int[b^{ax}]=\frac{b^{ax}}{aln(b)}\\&\text{Note for this problem, the }a\text{ value is } 20\\&\text{Taking the first integral we find:}\\&\frac{(17\cdot 2^{(20x)})}{(70ln(2))}+C_{1}\\&\text{Note that after indefinite integration we add a constant, }C_1\\&\text{Integrating again, we'll add a new constant, while multiplying}\\&\text{our first one by }x:\\&\frac{(17\cdot 2^{(20x)})}{(1400ln(2)^{2})}+C_{1}x+C_{2}\\&\end{align*}\)

\(\displaystyle \begin{align*}&\text{In performing an integral, familiarity with derivative rules}\\&\text{can prove useful. After all, an integral is essentially the}\\&\text{opposite operation of a derivative. Consider how in a derivative}\\&\text{we’d use the chain rule to multiply the derivative of the “inside”}\\&\text{of a function by the derivative of the outside? Integrals, in}\\&\text{similar fashion, entail a division.}\\&\text{Utilizing integral rules:}\\&\int[sin(ax)]=-\frac{cos(ax)}{a}\\&\int[cos(ax)]=\frac{sin(ax)}{a}\\&\text{Note for this problem, the }a\text{ value is } 5\\&\text{Taking the first integral we find:}\\&-\frac{(54cos(5x))}{35}+C_{1}\\&\text{Note that after indefinite integration we add a constant, }C_1\\&\text{Integrating again, we'll add a new constant, while multiplying}\\&\text{our first one by }x:\\&-\frac{(54sin(5x))}{175}+C_{1}x+C_{2}\\&\end{align*}\)

\(\displaystyle \begin{align*}&\text{In performing an integral, familiarity with derivative rules}\\&\text{can prove useful. After all, an integral is essentially the}\\&\text{opposite operation of a derivative. Consider how in a derivative}\\&\text{we’d use the chain rule to multiply the derivative of the “inside”}\\&\text{of a function by the derivative of the outside? Integrals, in}\\&\text{similar fashion, entail a division.}\\&\text{Utilizing integral rules:}\\&\int x^{a}=\frac{x^{a+1}}{a+1}\\&\text{Note for this problem, the }a\text{ value is } -18\\&\text{Taking the first integral we find:}\\&-\frac{3}{(1360x^{17})}+C_{1}\\&\text{Note that after indefinite integration we add a constant, }C_1\\&\text{Integrating again, we'll add a new constant, while multiplying}\\&\text{our first one by }x:\\&\frac{3}{(21760x^{16})}+C_{1}x+C_{2}\\&\end{align*}\)

\(\displaystyle \begin{align*}&\text{In performing an integral, familiarity with derivative rules}\\&\text{can prove useful. After all, an integral is essentially the}\\&\text{opposite operation of a derivative. Consider how in a derivative}\\&\text{we’d use the chain rule to multiply the derivative of the “inside”}\\&\text{of a function by the derivative of the outside? Integrals, in}\\&\text{similar fashion, entail a division.}\\&\text{Utilizing integral rules:}\\&\int\frac{a}{x}=aln(x)=ln(x^{a})\\&\int aln(x)=a(x(ln(x)-1))\\&\text{Note for this problem, the }a\text{ value is } \frac{5}{64}\\&\text{Taking the first integral we find:}\\&\frac{(5ln(x))}{64}+C_{1}\\&\text{Note that after indefinite integration we add a constant, }C_1\\&\text{Integrating again, we'll add a new constant, while multiplying}\\&\text{our first one by }x:\\&\frac{(5x\cdot (ln(x) - 1))}{64}+C_{1}x+C_{2}\\&\end{align*}\)

\(\displaystyle \begin{align*}&\text{In performing an integral, familiarity with derivative rules}\\&\text{can prove useful. After all, an integral is essentially the}\\&\text{opposite operation of a derivative. Consider how in a derivative}\\&\text{we’d use the chain rule to multiply the derivative of the “inside”}\\&\text{of a function by the derivative of the outside? Integrals, in}\\&\text{similar fashion, entail a division.}\\&\text{Utilizing integral rules:}\\&\int[b^{ax}]=\frac{b^{ax}}{aln(b)}\\&\text{Note for this problem, the }a\text{ value is } 11\\&\text{Taking the first integral we find:}\\&\frac{(5\cdot 2^{(11x)})}{(572ln(2))}+C_{1}\\&\text{Note that after indefinite integration we add a constant, }C_1\\&\text{Integrating again, we'll add a new constant, while multiplying}\\&\text{our first one by }x:\\&\frac{(5\cdot 2^{(11x)})}{(6292ln(2)^{2})}+C_{1}x+C_{2}\\&\end{align*}\)

\(\displaystyle \small \int\frac{3x}{x^2+x-2}dx\)

First factor the denominator and then write the results as a sum of two fractions as follows, 

\(\displaystyle \small \small \small \frac{3x}{x^2+x-2}=\frac{3x}{(x-1)(x+2)}=\frac{A}{x-1}+\frac{B}{x+2}\)                   (1)

Here we have applied the reverse of the cross multiplication process. Now we carefully deduce the numerators \(\displaystyle \small A\) and \(\displaystyle \small B\). Cross multiply as follows: 

 \(\displaystyle \small \frac{A}{x-1}+\frac{B}{x+2}=\frac{A(x+2)+B(x-1)}{(x-1)(x+2)}\)                                (2)

Looking at Equation (1) and Equation (2) it's clear that we must find values of \(\displaystyle \small A\)and \(\displaystyle \small B\) such that:

 \(\displaystyle \small A(x+2)+B(x-1)=3x\)                                                     (3)

Note that Equation (3) is an identity, meaning it must be true for all values of \(\displaystyle \small x.\) Therefore, we can freely choose a value of \(\displaystyle \small x\) to eliminate either \(\displaystyle \small A\) or \(\displaystyle \small B\). Choosing \(\displaystyle x=1\) eliminates \(\displaystyle B\). 

 \(\displaystyle \small \small x =1 \; \; \; \; \Rightarrow\; \; \; \;A(3)=3 \; \; \; \; \Rightarrow \; \; \; \; A=1\)

\(\displaystyle \small \small x=-2 \; \; \; \; \Rightarrow \; \; \; \;B(-3)=-6\; \; \; \; \Rightarrow \; \; \; \; B = 2\)

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Alternatively, \(\displaystyle \small A\) and \(\displaystyle \small B\) could have been deduced by expanding the left side of Equation (3) and collecting like terms as follows, 

\(\displaystyle \small Ax+2A+Bx-B =3x\)                                                        

\(\displaystyle \small (A+B)x+2A-B =3x\)                                                        (4)

 Upon inspection of Equation (4) we see that the two following conditions for \(\displaystyle \small A\) and \(\displaystyle \small B\) are required, 

\(\displaystyle \small A+B = 3\)

\(\displaystyle \small 2A-B =0\)

Solve for \(\displaystyle \small A\) and \(\displaystyle \small B\) , this can be done quickly by adding the two equations above, 

\(\displaystyle \small 3A=3 \: \: \: \rightarrow \: \: \:A=1\)

Using this value, \(\displaystyle \small \small A=1\) gives \(\displaystyle \small B = 2\). 

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Summarizing our partial fraction decomposition: 

\(\displaystyle \small \small \small \frac{3x}{x^2+x-2}=\frac{A}{x-1}+\frac{B}{x+2} =\frac{1}{x-1}+\frac{2}{x+2}\)

Now we can use the partial fraction decomposition to compute the integral.  

 \(\displaystyle \small \small \small \int\frac{3x}{x^2+x-2}dx=\int\left( \frac{1}{x-1}+\frac{2}{x+2}\right )dx\)

We can apply the sum rule on the right side and recognize that each term is the derivative of a natural logarithm, 

\(\displaystyle \small \small \small \int \frac{1}{x-1}dx +\int\frac{2}{x+2}dx = \ln(x-1)+2\ln(x+2)+C\)

We could use the properties of logarithms to write the result as a single logarithm, but we will leave it in the form of a sum here, remembering the constant of integration \(\displaystyle C\). 

\(\displaystyle \small \small \int\frac{3x}{x^2+x-2}dx= \ln(x-1)+2\ln(x+2)+C\)

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Now we can check to make sure the result is correct by differentiating the result: 

\(\displaystyle \frac{d}{dx}\left[\ln(x-1)+2\ln(x+2)+C \right ]\)

\(\displaystyle =\frac{1}{x-1}+\frac{2}{x+2}+0\)

\(\displaystyle =\frac{x+2}{(x-1)(x+2)}+\frac{2(x-1)}{(x-1)(x+2)}\)

\(\displaystyle =\frac{x+2+2(x-1)}{(x-1)(x+2)}\)

\(\displaystyle =\frac{3x}{x^2+x-2}\)

\(\displaystyle \int 5^x \cos(x)dx\)

 First we can start by calling the integral \(\displaystyle I\) for convenience in a later step.

\(\displaystyle I = \int 5^x\cos(x)dx\)                                                                         (1)

Let's rewrite the exponential function \(\displaystyle 5^x\) using the inverse relationship between \(\displaystyle e^x\) and \(\displaystyle \ln(x)\), 

\(\displaystyle 5^x = e^{\ln(5^x )} = e^{x\ln(5)}\) 

So now Equation (1) can be written as: 

 \(\displaystyle I = \int e^{x\ln(5)}\cos(x)dx\)                                                           (2)

 To evaluate this integral we will use integration by parts. We start by carefully defining two new variables that will break our integral down into more manageable terms. 

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Setup I.B.P

\(\displaystyle \int udv = uv - \int vdu\)                                                            (3) 

 \(\displaystyle u\) and \(\displaystyle v\) are both functions of \(\displaystyle x\). 

Determining how to define the new variables \(\displaystyle u\) and \(\displaystyle dv\) can be a process of trial and error. Once you define \(\displaystyle u\) and \(\displaystyle dv\), you can find the other parts, \(\displaystyle v\) and \(\displaystyle du\). 

 \(\displaystyle u = e^{x\ln(5)}\: \: \: \: \: \rightarrow \: \: \: \: \:du =\ln(5)e^{5\ln(x)}dx\)

 \(\displaystyle dv = \cos(x)dx\: \: \: \: \: \: \rightarrow \: \: \: \: \: \:v = \sin(x)\)______________________________________________________________

Now we can look off Equation (3) to assemble the integral, then simplify: 

\(\displaystyle I = \int e^{x\ln(5)}\cos(x)dx =e^{x\ln(5)}\sin(x) - \int \sin(x)\left(\ln(5)e^{x\ln(5)}dx \right )\)

 \(\displaystyle I =e^{x\ln(5)}\sin(x) - \ln(5)\int \sin(x)e^{x\ln(5)}dx\)                              (4)

Notice that the second term on the right-side of Equation (4) has yet another instance where we are multiplying an exponential function by a trigonometric function, a situation very similar to the original integral Equation (2).

Now let's perform integration by parts on the second term in Equation (4), ignoring the \(\displaystyle \ln(5)\) in front of the integral for a moment.    

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Setup I.B.P Again for \(\displaystyle \int \sin(x)e^{x\ln(5)}dx\)

\(\displaystyle \int u_1dv_1 =u_1v_1-\int v_1 du_1\)   

\(\displaystyle u_1 = e^{x\ln(5)}\: \: \: \: \: \: \rightarrow \: \: \: \: \: \:du_1 = \ln(5)e^{5\ln(x)}dx\)

\(\displaystyle dv_1 = \sin(x)dx \: \: \: \: \: \: \rightarrow \: \: \: \: \: \:v_1 = -\cos(x)dx\) 

 ______________________________________________________________ 

  Fill everything in and simplify:  

\(\displaystyle \int e^{x\ln(5)}\sin(x)dx=e^{xln(5)}(-\cos(x)) -\int -cos(x)ln(5)e^{5\ln(x)}dx\)

  \(\displaystyle \int e^{x\ln(5)}\sin(x)dx=-e^{xln(5)}(\cos(x)) + \ln(5)\underset{This\: \: is\: \: I}{ \underbrace{\int cos(x)e^{5\ln(x)}dx }}\)

Note that the original integral \(\displaystyle I\) has appeared in the equation. 

\(\displaystyle \int e^{x\ln(5)}\sin(x)dx=-e^{xln(5)}(\cos(x)) + \ln(5)I\)                         (6)

Now let's substitute Equation (6) this into Equation (4).  

\(\displaystyle I =e^{x\ln(5)}\sin(x) - \ln(5)\left[- e^{xln(5)}\cos(x) + \ln(5)I \right ]\)                   (7)

Now solve Equation (7) for \(\displaystyle I\). 

 \(\displaystyle I +[\ln(5)]^2 I=e^{x\ln(5)}\sin(x) + \ln(5)\-e^{xln(5)}\cos(x)\)

 \(\displaystyle I = \frac{e^{x\ln(5)}\sin(x) + \ln(5)\-e^{xln(5)}\cos(x)}{1+(\ln 5)^2}\)

Now we can write it using \(\displaystyle 5^x = e^{\ln(5^x )}\), 

 \(\displaystyle I = \frac{5^x\sin(x) + \ln(5)\ 5^x \cos(x)}{1+(\ln 5)^2}\)

Finally, 

 \(\displaystyle \int 5^x \cos(x)dx=\frac{5^x\left[\sin(x) + \ln(5)\cos(x) \right ]}{1+(\ln 5)^2}\)

To solve this equation we must integrate by parts by using the equation

\(\displaystyle \int udv=uv-\int vdu\).  

The first step in integrating by parts is to define u and dv.  

In this problem \(\displaystyle u=x\) and \(\displaystyle dv=e^{-x}dx\).  

We then need to find du and v.  

Taking the derivative of u we get \(\displaystyle du=1*dx\) and taking the integral of dv we get \(\displaystyle v=-e^{-x}\).  

Thus we can plug in the values and get 

\(\displaystyle \int xe{^{-x}}dx=-xe^{-x}-\int-e^{-x}dx=-x^{-x}-e^{-x}+C\)