Antiderivatives practice questions — AP Calculus BC

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

$$\int_{0}^{2} \frac{x}{(x^{2}+1)^{2}}\,dx = \frac{1}{2}\int_{1}^{5} u^{-2}\,du$$

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

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

Let $u=x$ and $\dfrac{\mathrm{d}v}{\mathrm{d}x}=e^{2x}$, so $\dfrac{\mathrm{d}u}{\mathrm{d}x}=1$ and $v=\tfrac12 e^{2x}$.

By parts, $\displaystyle\int u\,\mathrm{d}v = uv-\int v\,\mathrm{d}u$: $$\int x\,e^{2x}\,\mathrm{d}x = \tfrac12 x\,e^{2x}-\int \tfrac12 e^{2x}\,\mathrm{d}x.$$

The remaining integral gives $\tfrac14 e^{2x}$, so the antiderivative is $$\tfrac12 x\,e^{2x}-\tfrac14 e^{2x}.$$

Evaluate from $0$ to $1$: $$\left(\tfrac12 e^{2}-\tfrac14 e^{2}\right)-\left(0-\tfrac14\right)=\tfrac14 e^{2}+\tfrac14.$$

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

Take $u = x,\ dv = e^{x}\,dx$, giving $du = dx,\ v = e^{x}$: $$\int_0^1 x e^{x}\,dx = \Big[x e^{x}\Big]_0^1 - \int_0^1 e^{x}\,dx$$ $$= \big(1\cdot e - 0\big) - \Big[e^{x}\Big]_0^1 = e - (e - 1) = 1$$

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

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

$\int \!\left(\dfrac{1}{x} - \dfrac{1}{x+1}\right) dx = \ln|x| - \ln|x + 1| + C = \ln\!\left|\dfrac{x}{x + 1}\right| + C$.

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

$$x:0\to\sqrt{\pi}\ \Longrightarrow\ u:0\to\pi$$

$$\int_{0}^{\sqrt{\pi}} x\,\sin\!\left(x^{2}\right)\,dx = \frac{1}{2}\int_{0}^{\pi}\sin u\,du$$

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

$\int \cos x \, dx = \sin x + C$.

$\int \dfrac{dx}{x} = \ln|x| + C$. For $x > 0$: $\ln x + C$.