Antiderivatives

$\int cdx=cx+C$. So $\int 5dx=5x+C$.

The exponential is its own antiderivative: $\int e^{x}dx=e^x+C$.

$\int x^{3}dx={\displaystyle \frac{x^4}{4}}+C$.

Let $u=\sin x$, $du=\cos xdx$. The integral becomes ${\int }_0^1{u}^{3}du={\left[{\displaystyle \frac{u^4}{4}}\right]}_0^1={\displaystyle \frac{1}{4}}$.

$\int {\displaystyle \frac{dx}{x}}=\ln \mathrm{\mid }x\mathrm{\mid }+C$. For $x>0$: $\ln x+C$.

$\int x{e}^{x}dx=x{e}^x-\int e^{x}dx=x{e}^x-e^x+C=(x-1)e^x+C$.

A standard antiderivative: $\int {\displaystyle \frac{dx}{1+x^2}}=\arctan x+C$.

$u=x^2\Rightarrow du=2xdx$. So $\int \cos udu=\sin u+C=\sin (x^2)+C$.

$\int \cos xdx=\sin x+C$.

$\int x{e}^{x}dx=x{e}^x-\int e^{x}dx=(x-1)e^x+C$. Evaluating from $0$ to $1$: $(0)\cdot e-(-1)\cdot 1=0-(-1)=1$.