\(\displaystyle \int x\sqrt{1+x^2}dx\)

Integrals such as this are seen very commonly in introductory calculus courses. It is often useful to look for patterns such as the fact that the polynomial under the radical in our example, \(\displaystyle 1+x^2\), happens to be one order higher than the factor outside the radical, \(\displaystyle x.\) You know that if you take a derivative of a second order polynomial you will get a first order polynomial, so let's define the variable: 

\(\displaystyle u = 1+x^2\)                                                            (1)

Now differentiate with respect to \(\displaystyle x\) to write the differential for \(\displaystyle u\), 

\(\displaystyle du = 2x dx\)                                                            (2)

Looking at equation (2), we can solve for \(\displaystyle xdx\), to obtain  \(\displaystyle xdx =\frac{1}{2}du\). Now if we look at the original integral we can rewrite in terms of \(\displaystyle u\)

\(\displaystyle \int x\sqrt{1+x^2}dx = \int\sqrt{1+x^2}\underbrace{xdx} = \int\sqrt{u}\left(\frac{1}{2}du \right )\)                    

Now proceed with the integration with respect to \(\displaystyle u\). 

\(\displaystyle \int\sqrt{u}\left(\frac{1}{2}du \right ) = \frac{1}{2}\int u^{\frac{1}{2}}du\) 

\(\displaystyle = \frac{1}{2}\left[\frac{1}{\frac{1}{2}+1}u^{\frac{1}{2}+1} \right ]+C\)

\(\displaystyle =\frac{1}{2} \left(\frac{2}{3}u^{\frac{3}{2}} \right )+C\)

\(\displaystyle =\frac{1}{2} \left(\frac{2}{3}u^{\frac{3}{2}} \right )+C\)

\(\displaystyle =\frac{1}{3}u^{\frac{3}{2}}+C\)

Now write the result in terms of \(\displaystyle x\) using equation (1), we conclude,  

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

Let \(\displaystyle u=sin(x)\)

Then \(\displaystyle du=cos(x)dx\)

Now we can subtitute

\(\displaystyle \int sin^2(x) \cos(x) dx=\int u^2du= \frac{1}{3}u^3\)

Now we substitute back

\(\displaystyle \frac{1}{3}u^3=\frac{1}{3} \sin^3(x)\)

We can use substitution for this integral.

Let \(\displaystyle u = 1+x^2\),

then \(\displaystyle du = 2x dx\).

Multiplying this last equation by \(\displaystyle 3\), we get \(\displaystyle 3du = 6x dx\).

Now we can make our substitutions

\(\displaystyle \int_0^1 \frac{6x}{1+x^2} dx\). Start

\(\displaystyle =\int_1^2 \frac{3}{u}du\). Swap out \(\displaystyle 1+x^2\) with \(\displaystyle u\), and \(\displaystyle 6x dx\) with \(\displaystyle 3du\). Make sure you also plug the bounds on the integral into \(\displaystyle u = 1+x^2\) for \(\displaystyle x\) to get the new bounds.

\(\displaystyle =3\int_1^2 \frac{1}{u}du\). Factor out the \(\displaystyle 3\).

\(\displaystyle =3 \ln(u)|_1^2\). Integrate (absolute value signs are not needed since \(\displaystyle 1\le u \le2\).)

\(\displaystyle =3(\ln(2)-\ln(1))\). Evaluate

\(\displaystyle =3\ln2 -0 = 3\ln2\).

To solve the integral, we have to simplify it by using a variable u to substitute for a variable of x.

For this problem, we will let u replace the expression \(\displaystyle 2+x^4\).

Next, we must take the derivative of u. Its derivative is \(\displaystyle du=4x^3dx\).

Next, solve this equation for dx so that we may replace it in the integral.

Plug \(\displaystyle u\) in place of \(\displaystyle (2+x^4)\)and \(\displaystyle \frac{du}{4x^3}\) in place of \(\displaystyle dx\) into the original integral and simplify.

The \(\displaystyle x^3\) in the denominator cancels out the remaining \(\displaystyle x^3\) in the integral, leaving behind a \(\displaystyle \frac{1}{4}\). We can pull the \(\displaystyle \frac{1}{4}\) out front of the integral. Next, take the anti-derivative of the integrand and replace u with the original expression, adding the constant \(\displaystyle c\) to the answer.  

The specific steps are as follows:

1. \(\displaystyle \int{x^3(2+x^4)^5dx}\)

2.\(\displaystyle u=2+x^4\)

3. \(\displaystyle du=4x^3dx\)

4. \(\displaystyle dx=\frac{du}{4x^3}\)

5. \(\displaystyle \int{x^3u^5(\frac{du}{4x^3})}\)

6. \(\displaystyle \frac{1}{4}\int{u^5}du\)

7. \(\displaystyle \frac{1}{4}(\frac{1}{6}u^6)\)

8. \(\displaystyle \frac{1}{4}(\frac{1}{6}(2+x^4)^6\)

9. \(\displaystyle \frac{1}{24}(2+x^4)^6+c\)

To solve the integral, we have to simplify it by using a variable \(\displaystyle u\) to substitute for a variable of \(\displaystyle x\). For this problem, we will let u replace the expression \(\displaystyle x^3+1\). Next, we must take the derivative of u. Its derivative is \(\displaystyle 3x^2dx\). Next, solve this equation for \(\displaystyle dx\) so that we may replace it in the integral. Plug \(\displaystyle u\) in place of \(\displaystyle x^3+1\) and \(\displaystyle \frac{du}{3x^2}\) in place of \(\displaystyle dx\) into the original integral and simplify. The \(\displaystyle x^2\) in the denominator cancels out the remaining \(\displaystyle x^2\) in the integral, leaving behind a \(\displaystyle \frac{1}{3}\). We can pull the \(\displaystyle \frac{1}{3}\) out front of the integral. Next, take the anti-derivative of the integrand and replace \(\displaystyle u\) with the original expression, adding the constant \(\displaystyle c\) to the answer. The specific steps are as follows:

1. \(\displaystyle \int{x^2\sqrt{x^3+1}}dx\)

2. \(\displaystyle u\)=\(\displaystyle x^3+1\)

3. \(\displaystyle du=3x^2dx\)

4.\(\displaystyle \int{x^2\sqrt{u}\frac{du}{3x^2}}\)

5. \(\displaystyle \frac{1}{3}\int{\sqrt{u}du}\)

6. \(\displaystyle \frac{1}{3}(\frac{2}{3}u^{\frac{3}{2}})+c\)

7. \(\displaystyle \frac{2}{9}u^{\frac{3}{2}}+c\)

8. \(\displaystyle \frac{2}{9}(x^3+1)^{\frac{3}{2}}+c\)

To solve the integral, we have to simplify it by using a variable \(\displaystyle u\) to substitute for a variable of \(\displaystyle x\). For this problem, we will let u replace the expression \(\displaystyle cosx\). Next, we must take the derivative of \(\displaystyle u\). Its derivative is \(\displaystyle -sinxdx\).  Next, solve this equation for \(\displaystyle dx\) so that we may replace it in the integral. Plug \(\displaystyle u\) in place of \(\displaystyle cosx\) and \(\displaystyle \frac{du}{-sinx}\) in place of \(\displaystyle dx\) into the original integral and simplify. The \(\displaystyle sinx\) in the denominator cancels out the remaining \(\displaystyle sinx\) in the integral, leaving behind a \(\displaystyle -1\). Next, take the anti-derivative of the integrand and replace \(\displaystyle u\) with the original expression, adding the constant \(\displaystyle c\) to the answer. The specific steps are as follows:

1. \(\displaystyle \int{cos^3xsinxdx}\)

2. \(\displaystyle u=cosx\)

3. \(\displaystyle du=-sinxdx\)

4. \(\displaystyle \int{u^3sinx(\frac{du}{-sinx})}\)

5. \(\displaystyle \int-u^3du\)

6. \(\displaystyle \frac{-1}{4}u^4+c\)

7. \(\displaystyle \frac{-cos^4x}{4}+c\)

To solve the integral, we have to simplify it by using a variable \(\displaystyle u\) to substitute for a variable of \(\displaystyle x\). For this problem, we will let u replace the expression \(\displaystyle x^2\). Next, we must take the derivative of \(\displaystyle u\). Its derivative is \(\displaystyle 2x\). Next, solve this equation for \(\displaystyle dx\) so that we may replace it in the integral. Plug \(\displaystyle u\) in place of \(\displaystyle x^2\) and \(\displaystyle \frac{du}{2x}\) in place of \(\displaystyle dx\) into the original integral and simplify. The \(\displaystyle x\) in the denominator cancels out the remaining \(\displaystyle x\) in the integral, leaving behind a \(\displaystyle \frac{1}{2}\).  We can pull the \(\displaystyle \frac{1}{2}\) out front of the integral. Next, take the anti-derivative of the integrand and replace \(\displaystyle u\) with the original expression, adding the constant \(\displaystyle c\) to the answer. The specific steps are as follows:

1. \(\displaystyle \int{xsin(x^2)dx}\)

2. \(\displaystyle u=x^2\)

3. \(\displaystyle du=2xdx\)

4. \(\displaystyle \int{x(sinu)\frac{du}{2x}}\)

5. \(\displaystyle \frac{1}{2}\int{sinudu}\)

6. \(\displaystyle \frac{-1}{2}cosu+c\)

7. \(\displaystyle \frac{-1}{2}cos(x^2)+c\)

To solve the integral, we have to simplify it by using a variable \(\displaystyle u\) to substitute for a variable of \(\displaystyle x\). For this problem, we will let u replace the expression \(\displaystyle x^3\). Next, we must take the derivative of \(\displaystyle u\). Its derivative is \(\displaystyle 3x^2\). Next, solve this equation for \(\displaystyle dx\) so that we may replace it in the integral. Plug \(\displaystyle u\) in place of \(\displaystyle x^3\) and \(\displaystyle \frac{du}{3x^2}\) in place of \(\displaystyle dx\) into the original integral and simplify. The \(\displaystyle x^2\) in the denominator cancels out the remaining \(\displaystyle x^2\) in the integral, leaving behind a \(\displaystyle \frac{1}{3}\). We can pull the \(\displaystyle \frac{1}{3}\) out front of the integral. Next, take the anti-derivative of the integrand and replace \(\displaystyle u\) with the original expression, adding the constant \(\displaystyle c\) to the answer. The specific steps are as follows:

1. \(\displaystyle \int{x^2e^{x^3}dx}\)

2. \(\displaystyle u=x^3\)

3. \(\displaystyle du=3x^2\)

4. \(\displaystyle \int{x^2e^u(\frac{du}{3x^2})}\)

5. \(\displaystyle \frac{1}{3}\int{e^udu}\)

6. \(\displaystyle \frac{1}{3}e^u+c\)

7. \(\displaystyle \frac{1}{3}e^{x^3}+c\)

To solve the integral, we have to simplify it by using a variable \(\displaystyle u\) to substitute for a variable of \(\displaystyle x\). For this problem, we will let u replace the expression \(\displaystyle e^x\). Next, we must take the derivative of \(\displaystyle u\). Its derivative is \(\displaystyle e^x\). Next, solve this equation for \(\displaystyle dx\) so that we may replace it in the integral. Plug \(\displaystyle u\) in place of \(\displaystyle e^x\) and \(\displaystyle \frac{du}{e^x}\) in place of \(\displaystyle dx\) into the original integral and simplify. The \(\displaystyle e^x\) in the denominator cancels out the remaining \(\displaystyle e^x\) in the integral. Next, take the anti-derivative of the integrand and replace \(\displaystyle u\) with the original expression, adding the constant \(\displaystyle c\) to the answer. The specific steps are as follows:

1. \(\displaystyle \int{e^xcos(e^x)dx}\)

2. \(\displaystyle u=e^x\)

3. \(\displaystyle du=e^xdx\)

4. \(\displaystyle \int{e^xcosu(\frac{du}{e^x})}\)

5. \(\displaystyle \int{cosudu}\)

6. \(\displaystyle sinu+c\)

7. \(\displaystyle sin(e^x)+c\)

To solve the integral, we have to simplify it by using a variable \(\displaystyle u\) to substitute for a variable of \(\displaystyle x\). For this problem, we will let u replace the expression \(\displaystyle lnx\). Next, we must take the derivative of \(\displaystyle u\). Its derivative is \(\displaystyle \frac{1}{x}\). Next, solve this equation for \(\displaystyle dx\) so that we may replace it in the integral. Plug \(\displaystyle u\) in place of \(\displaystyle lnx\) and \(\displaystyle dux\) in place of \(\displaystyle dx\) into the original integral and simplify. The \(\displaystyle x\) in the denominator cancels out the remaining \(\displaystyle x\) in the integral. Next, take the anti-derivative of the integrand and replace \(\displaystyle u\) with the original expression, adding the constant \(\displaystyle c\) to the answer. The specific steps are as follows:

1. \(\displaystyle \int{\frac{sin(lnx)}{x}dx}\)

2. \(\displaystyle u=lnx\)

3. \(\displaystyle du=(\frac{1}{x}dx)\)

4. \(\displaystyle dux=dx\)

5. \(\displaystyle \int{\frac{sinu}{x}dux}\)

6. \(\displaystyle \int{sinudu}\)

7. \(\displaystyle -cosu+c\)

8. \(\displaystyle -cos(lnx)+c\)