11 questions · self-marked practice — reveal the answer and mark yourself.
Problem:
$\int \frac{\log x}{(1+\log x)^2} d x$
adding and substracting 1 from numerator
$\begin{aligned} & \int \frac{1-1+\log x}{(1+\log x)^2} d x \\ & \int \frac{1+\log x}{(1+\log x)^2} d x-\int \frac{1}{(1+\log x)^2} d x \\ & \int \frac{1}{1+\log x} d x-\int \frac{1}{(1+\log x)^2} d x\end{aligned}$
For the integral
$\int \frac{1}{1+\log x} d x$
integrate by parts within the sum: ∫fg'=fg−∫f'g
$\begin{aligned} & f=\frac{1}{1+\log x} d x, g \prime=1 \\ & f \prime=-\frac{1}{(1+\log x)^2}, g=x \\ & =-\int \frac{1}{(1+\log x)^2} d x-\int-\frac{1}{(1+\log x)^2} d x+\frac{x}{\log (x)+1} \\ & =\frac{x}{\log (x)+1}\end{aligned}$
$\begin{aligned} & \int \frac{1}{a^2-x^2} d x=\int \frac{1}{(a-x)(a+x)} d x \\ & =\frac{1}{2 a} \int \frac{(a-x)+(a+x)}{(a-x)(a+x)} d x \\ & =\frac{1}{2 a} \int\left(\frac{1}{a+x}+\frac{1}{a-x}\right) d x \\ & =\frac{1}{2 a}\left[\int \frac{1}{a+x} d x+\int \frac{1}{a-x} d x\right]\end{aligned}$
$\begin{aligned} & =\frac{1}{2 a}\left[\log |a+x|+\frac{\log |a-x|}{-1}\right]+C=\frac{1}{2 a}[\log |a+x|-\log |a-x|]+C \\ & =\frac{1}{2 a} \log \left|\frac{a+x}{a-x}\right|+C\end{aligned}$
$\begin{aligned} & \text { Let } I =\int \sqrt{a^2-x^2} d x \\ & =\int \sqrt{a^2-x^2} \cdot 1 d x \\ & =\sqrt{a^2-x^2} \cdot \int 1 d x-\int\left[\frac{d}{d x}\left(\sqrt{a^2-x^2}\right) \cdot \int 1 d x\right] d x \\ & =\sqrt{a^2-x^2} \cdot x-\int\left[\frac{1}{2 \sqrt{a^2-x^2}} \cdot \frac{d}{d x}\left(a^2-x^2\right) \cdot x\right] d x \\ & =\sqrt{a^2-x^2} \cdot x-\int \frac{1}{2 \sqrt{a^2-x^2}}(0-2 x) \cdot x d x \\ & =\sqrt{a^2-x^2} \cdot x-\int \frac{-x}{\sqrt{a^2-x^2}} \cdot x d x \\ & =x \sqrt{a^2-x^2}-\int \frac{a^2-x^2-a^2}{\sqrt{a^2-x^2}} d x \\ & =x \sqrt{a^2-x^2}-\int \sqrt{a^2-x^2} d x+a^2 \int \frac{d x}{\sqrt{a^2-x^2}}\end{aligned}$
$\begin{aligned} & =x \sqrt{a^2-x^2}-I+a^2 \sin ^{-1}\left(\frac{x}{a}\right)+c_1 \\ & \therefore 2 I =x \sqrt{a^2-x^2}+a^2 \sin ^{-1}\left(\frac{x}{a}\right)+c_1\end{aligned}$$\begin{aligned} & \therefore I =\frac{x}{2} \sqrt{a^2-x^2}+\frac{a^2}{2} \sin ^{-1}\left(\frac{x}{a}\right)+\frac{c_1}{2} \\ & \therefore \int \sqrt{a^2-x^2} d x=\frac{x}{2} \sqrt{a^2-x^2}+\frac{a^2}{2} \sin ^{-1}\left(\frac{x}{a}\right)+c_1\end{aligned}$where $c =\frac{c_1}{2}$
$\begin{aligned} & \text { Let } I =\int \sqrt{x^2+a^2} d x \\ & =\int \sqrt{x^2+a^2} \cdot 1 d x\end{aligned}$$\begin{aligned} & =\sqrt{x^2+a^2} \int 1 d x-\int\left[\frac{d}{d x}\left(\sqrt{x^2+a^2}\right) \cdot \int 1 d x\right] d x \\ & =\sqrt{x^2+a^2} \cdot x-\int \frac{2 x}{2 \sqrt{x^2+a^2}} \cdot x d x \\ & =x \cdot \sqrt{x^2+a^2}-\int \frac{\left(x^2+a^2\right)-a^2}{\sqrt{x^2+a^2}} d x\end{aligned}$$\begin{aligned} & =x \cdot \sqrt{x^2+a^2}-\int\left(\frac{x^2+a^2}{\sqrt{x^2+a^2}}-\frac{a^2}{\sqrt{x^2+a^2}}\right) d x \\ & =x \cdot \sqrt{x^2+a^2}-\int \sqrt{x^2+a^2} d x+a^2 \int \frac{1}{\sqrt{x^2+a^2}} d x \\ & \therefore I =x \cdot \sqrt{x^2+a^2}-I+a^2 \log \left|x+\sqrt{x^2+a^2}\right|+c_1 \\ & \therefore 2 I =x \cdot \sqrt{x^2+a^2}+a^2 \log \left|x+\sqrt{x^2+a^2}\right|+c_1 \\ & \therefore I=\frac{x}{2} \sqrt{x^2+a^2}+\frac{a^2}{2} \log \left|x+\sqrt{x^2+a^2}\right|+\frac{c_1}{2}\end{aligned}$$\therefore \int \sqrt{x^2+a^2} d x=\frac{x}{2} \sqrt{x^2+a^2}+\frac{a^2}{2} \log \left|x+\sqrt{x^2+a^2}\right|+c$, where $c=\frac{c_1}{2}$