Antiderivatives practice questions — AP Calculus BC

Substitute $u=x^2+1$, giving $du=2xdx$, so $xdx=\frac{1}{2}du$. The limits become $x=0\Rightarrow u=1$ and $x=2\Rightarrow u=5$:

$${\int }_0^2\frac{x}{(x^2+1{)}^2}dx=\frac{1}{2}{\int }_1^5{u}^{-2}du$$

$$\frac{1}{2}{[-\frac{1}{u}]}_1^5=\frac{1}{2}(1-\frac{1}{5})=\frac{2}{5}$$

$u=x$, $du=dx$, $v=\sin x$. $\int x\cos xdx=x\sin x-\int \sin xdx=x\sin x+\cos x+C$.

Let $u=x$ and ${\displaystyle \frac{\mathrm{d}v}{\mathrm{d}x}}=e^{2x}$, so ${\displaystyle \frac{\mathrm{d}u}{\mathrm{d}x}}=1$ and $v=\frac{1}{2}{e}^{2x}$.

By parts, ${\displaystyle \int u\mathrm{d}v=uv-\int v\mathrm{d}u}$: $$\int x{e}^{2x}\mathrm{d}x=\frac{1}{2}x{e}^{2x}-\int \frac{1}{2}{e}^{2x}\mathrm{d}x\mathrm{.}$$

The remaining integral gives $\frac{1}{4}{e}^{2x}$, so the antiderivative is $$\frac{1}{2}x{e}^{2x}-\frac{1}{4}{e}^{2x}\mathrm{.}$$

Evaluate from $0$ to $1$: $$(\frac{1}{2}{e}^2-\frac{1}{4}{e}^2)-(0-\frac{1}{4})=\frac{1}{4}{e}^2+\frac{1}{4}\mathrm{.}$$

Hence the value is $$\frac{e^2+1}{4}\mathrm{.}$$

Take $u=x,dv=e^{x}dx$, giving $du=dx,v=e^x$: $${\int }_0^{1}x{e}^{x}dx=[x{e}^x{]}_0^1-{\int }_0^1{e}^{x}dx$$ $$=(1\cdot e-0)-[e^x{]}_0^1=e-(e-1)=1$$

Apply integration by parts with $u=x$, $dv=\cos xdx$, so $du=dx$, $v=\sin x$: $$\int x\cos xdx=x\sin x-\int \sin xdx$$ $$\int x\cos xdx=x\sin x+\cos x+C$$ Check by differentiating: ${\displaystyle \frac{d}{dx}}(x\sin x+\cos x)=\sin x+x\cos x-\sin x=x\cos x$.

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

$\int \text{\hspace{0.17em}⁣}({\displaystyle \frac{1}{x}}-{\displaystyle \frac{1}{x+1}})dx=\ln \mathrm{\mid }x\mathrm{\mid }-\ln \mathrm{\mid }x+1\mathrm{\mid }+C=\ln \text{\hspace{0.17em}⁣}\mid {\displaystyle \frac{x}{x+1}}\mid +C$.

Substitute $u=x^2$, giving $du=2xdx$, so $xdx=\frac{1}{2}du$.

$$x:0\to \sqrt{\pi }⟹u:0\to \pi$$

$${\int }_0^{\sqrt{\pi }}x\sin \text{\hspace{0.17em}⁣}\left(x^2\right)dx=\frac{1}{2}{\int }_0^{\pi }\sin udu$$

$$=\frac{1}{2}[-\cos u{]}_0^{\pi }=\frac{1}{2}(-\cos \pi +\cos 0)=\frac{1}{2}(1+1)=1$$

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

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