Notice that $2x$‍ is the derivative of $x^2$‍. If we let $u=x^2$‍, then $du=2xdx$‍, and we can substitute to obtain a simpler integral.

We can now find the antiderivative:

So ${\displaystyle \int \cos \left(x^2\right)2xdx=\sin (x^2)+C}$‍

We can differentiate $\sin (x^2)+C$‍ to check our work.

Notice that $3x^2$‍ is the derivative of $x^3+3$‍. If we let $u=x^3+3$‍, then $du=3x^{2}dx$‍, and we can substitute to obtain a simpler integral.

We can now find the antiderivative:

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

We can differentiate $-{\displaystyle \frac{1}{x^3+3}}+C$‍ to check our work.

If we let $u=4x$‍, then $du=4dx$‍ and $dx={\displaystyle \frac{du}{4}}$‍.

Substituting, the integral becomes:

We can now find the antiderivative:

So ${\displaystyle \int e^{4x}dx={\displaystyle \frac{1}{4}}{e}^{4x}+C}$‍ .

If we let $u={\displaystyle \frac{1}{6}}{x}^2+1$‍, then $du={\displaystyle \frac{1}{3}}xdx$‍ and $xdx=3du$‍.

Substituting, the integral becomes:

We can now find the antiderivative:

So ${\displaystyle \int x\cdot \sqrt{{\displaystyle \frac{1}{6}}{x}^2+1}dx=2{({\displaystyle \frac{1}{6}}{x}^2+1)}^{\frac{3}{2}}+C}$‍ .