What is the
integral int (x^2+1)(x^3+3x)^4 dx?

$\displaystyle\int{\left({{x}}^{{2}}+{1}\right)}{{\left({{x}}^{{3}}+{3}{x}\right)}}^{{4}}{\left.{d}{x}\right.}$

Use U-substitution to solve this.

$\displaystyle{U}={{x}}^{{3}}+{3}{x}$

Take the derivative. Notice that there is a common factor of three!

$\displaystyle{d}{U}={3}{{x}}^{{2}}+{3}{\left.{d}{x}\right.}={3}{\left({{x}}^{{2}}+{1}\right)}{\left.{d}{x}\right.}$

Divide by three on both sides, and
then we will see that we can plug in $\displaystyle{\left({{x}}^{{2}}+{1}\right)}{\left.{d}{x}\right.}$

$\displaystyle\frac{{1}}{{3}}{d}{U}={\left({{x}}^{{2}}+{1}\right)}$
dx

Now that we know and , substitute these back to the integral.

$\displaystyle\int\ldots$

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Notes