This is a u-substitution integral.  We need to substitute the new function, which is modifying our base function (the exponential).

\(\displaystyle \int e^xdx=e^x+C\), but instead of that, our problem is \(\displaystyle e^u\).  We can solve this integral by completing the substitution.

\(\displaystyle u=4x-2.\; du=4dx. \;dx=\frac{du}{4}.\)

Now, we can replace everything in our integrand and rewrite in terms of our new variables:

\(\displaystyle \int e^{4x-2}dx=\int e^u\frac{du}{4}=\frac{1}{4}\int e^udu.\)

\(\displaystyle =\frac{1}{4}e^u+C=\frac{1}{4}e^{4x-2}+C\).

Remember to plug your variable back in and include the integration constant since we have an indefinite integral.  

This is a u-substitution integral.  We need to choose the following substitutions:

\(\displaystyle u=2x.\; du=2dx.\)

Now, we can replace our original problem with our new variables:

\(\displaystyle \int2cos(2x)dx=\int cos(2x)2dx=\int cos(u)du\)

\(\displaystyle =sin(u)+C=sin(2x)+C.\)

In the last step, we need to plug in our original function and add the integration constant. 

This is a hidden u-substitution problem!  Because we have a function under our square root, we cannot just simply integrate it.  Therefore, we need to choose the function under the square root as our substitution variable!

\(\displaystyle u=3+4x.\; du=4dx.\; dx=\frac{du}{4}.\)

Now, let us rewrite our original equation in terms of our new variable!

\(\displaystyle \int \sqrt{3+4x}dx=\int \sqrt{u}\frac{du}{4}=\frac{1}{4}\int \sqrt udu.\)

\(\displaystyle =\frac{1}{4}\int u^{1/2}=\frac{1}{4}\frac{u^{3/2}}{3/2}+C=\frac{1}{4}*\frac{2}{3}u^{3/2}+C\)

\(\displaystyle =\frac{2}{12}u^{3/2}+C=\frac{1}{6}({3+4x)}^{3/2}+C\).

This is a u-substitution problem.  We need to find a function and its derivative in the integral.

\(\displaystyle u=x^2-x+8.\; du=(2x-1)dx.\)

Now, replace your variables, and integrate.

\(\displaystyle \int (2x-1)(x^2-x+8)^7dx=\int (x^2-x+8)^7(2x-1)dx\)

\(\displaystyle =\int u^7du=\frac{u^8}{8}+C=\frac{(x^2-x+8)^8}{8}+C\).

This problem is an application of the u-substitution method.

\(\displaystyle u=3+x.\; x=u-3\; du=1dx.\)

Now, be careful that you replace everything in the original integral in terms of our new variables.  This includes the \(\displaystyle x\) term!

\(\displaystyle \int x\sqrt{3+x}dx=\int(u-3)u^{1/2}du=\int u^{3/2}-3u^{1/2}du\)

\(\displaystyle =\frac{u^{5/2}}{5/2}-3\frac{u^{3/2}}{3/2}=\frac{2}{5}u^{5/2}-2u^{3/2}+C\)

\(\displaystyle =\frac{2}{5}(3+x)^{5/2}-2(3+x)^{3/2}+C\).

To simplify the integral, we need to substitute new variables:

\(\displaystyle \\u=2x^2-8x+3.\; \\du=(4x-8)dx \\du=2(2x-4)dx \\\frac{1}{2}du=(2x-4)dx\)

Now, we can replace our original variables, and integrate!

\(\displaystyle \int \frac{2x-4}{2x^2-8x+3}dx=\int \frac{du}{2u}=\frac{1}{2}\int\frac{du}{u}\)

\(\displaystyle \\=\frac{1}{2}ln|u|+C \\\\ =\frac{1}{2}ln{|2x^2-8x+3|}+C\).

This is a hidden u-substitution problem!  Remember, to use substitution, we need to have an integral where a function and its derivative live inside.  If you look closely, you will see we have just that!

\(\displaystyle u=lnx.\; du=\frac{1}{x}dx.\)

Now, rewrite the integral, and integrate:

\(\displaystyle \int \frac{2lnx}{x}==2\int lnx*\frac{1}{x}dx=2\int u du=2\frac{u^2}{2}+C\)

\(\displaystyle =u^2+C=(lnx)^2+C.\)

To integrate, we must make the following substitution:

\(\displaystyle u=y^3, du=3y^2dy\)

The derivative was found using the following rule:

\(\displaystyle \frac{\mathrm{d} }{\mathrm{d} x} x^n=nx^{n-1}\)

Now, we rewrite the integral in terms of u and solve:

\(\displaystyle \frac{5}{3}\int\cos(u)du=\frac{5}{3}\sin(u)+C\)

The integral was found using the following rule:

\(\displaystyle \int \cos(x)dx=\sin(x)+C\)

Finally, replace u with our original x term:

\(\displaystyle \frac{5}{3}\sin(y^3)+C\)

To integrate, we must perform the following substitution:

\(\displaystyle u=x^2, du=2xdx\)

The derivative was found using the following rule:

\(\displaystyle \frac{\mathrm{d} }{\mathrm{d} x} x^n=nx^{n-1}\)

Now, we rewrite the integral in terms of u and solve:

\(\displaystyle \frac{3}{2}\int\ e^udu=\frac{3}{2}e^u+C\)

The integral was found using the following rule:

\(\displaystyle \int e^xdx =e^x+C\)

Finally, replace u with our original x term:

\(\displaystyle \frac{3}{2}e^{x^2}+C\)

To integrate, the following substitution was made:

\(\displaystyle u=3x^5, du=15x^4dx\)

Now, we rewrite the integral in terms of u and integrate:

\(\displaystyle \frac{7}{15}\int\sin(u)du=-\frac{7}{15}\cos(u)+C\)

The following rule was used for integration:

\(\displaystyle \int \sin(x)dx=-\cos(x)+C\)

Finally, rewrite the final answer in terms of our original x term:

\(\displaystyle -\frac{7}{15}\sin(3x^5)+C\)