Maths Solve the Following Question.(3 Marks)

Let $I=\int \frac{1+\log x}{x(2+\log x)(3+\log x)} d x$
Put
$\log x=t$
$\frac{1}{x} d x=d t$
$I=\int \frac{1+t}{(2+t)(3+t)} d t$
consider
$\frac{1+t}{(2+t)(3+t)}=\frac{A}{2+t}+\frac{B}{3+t}$
$(1+t)=A(3+t)+B(2+t)$
A=-1,B=2
$\begin{aligned} & \frac{1+t}{(2+t)(3+t)}=-\frac{1}{2+t}+\frac{2}{3+t} \\ & I=\int-\frac{1}{2+t} d t+\int \frac{2}{3+t} d t \\ & =-\log |(2+t)|+2 \log |(3+t)|+c \\ & =\log \left[\left|\frac{(3+t)^2}{2+t}\right|\right]+c \\ & =\log \left[\left|\frac{(3+\log x)^2}{2+\log x}\right|\right]+C\end{aligned}$

$\begin{aligned} & I=\int \sec ^3 x d x \\ & I=\int \sec x \cdot \sec ^2 x d x \\ & I=\sec x \cdot \int \sec ^2 x d x-\int\left[\frac{d}{d x}(\sec x) \cdot \int \sec ^2 x d x\right] d x \\ & I=\sec x \cdot \tan x-\int \sec x \cdot \tan x \cdot \tan x d x \\ & I=\sec x \cdot \tan x-\int \sec x\left(\sec ^2 x-1\right) d x \\ & I=\sec x \cdot \tan x-\int\left[\sec ^3 x-\sec x\right] d x \\ & I=\sec x \cdot \tan x-\int \sec ^3 x+\int \sec x d x \\ & I=\sec x \cdot \tan x-I+\log |\sec x+\tan x|+c \\ & 2 I=\sec x \cdot \tan x+\log |\sec x+\tan x|+c \\ & \therefore I=\frac{1}{2}(\sec x \cdot \tan x+\log |\sec x+\tan x|)+c\end{aligned}$

Let $I=\int \frac{1}{3+5 \cos x} d x$ put $\tan \left(\frac{x}{2}\right)=t$
then $d x=\frac{2}{1+t^2} d t$ and $\cos x=\frac{1-t^2}{1+t^2}$
$I=\int \frac{2 \frac{d t}{1+t^2}}{3+5\left(\frac{1-t^2}{1+t^2}\right)}$
$=2 \int \frac{\frac{d t}{1+t^2}}{\frac{3\left(1+t^2\right)+5\left(1-t^2\right)}{1+t^2}}$
$=2 \int \frac{d t}{3+3 t^2+5-5 t^2}$
$=2 \int \frac{d t}{8-2 t^2}$
$=\int \frac{d t}{2^2-t^2}$
$\begin{aligned} & =\frac{1}{2(2)} \log \left|\frac{2+t}{2-t}\right|+c \\ & =\frac{1}{4} \log \left|\frac{2+\tan \left(\frac{x}{2}\right)}{2-\tan \left(\frac{x}{2}\right)}\right|+c\end{aligned}$

Let $\int v d x=w \ldots . .(1)$
then $\frac{d w}{d x}=v \ldots . .(2)$
$N o w \frac{d}{d x}(u, w)=u \cdot \frac{d}{d x}(w)+w \frac{d}{d x}(u)$
$=u . v+w \frac{d u}{d x} \ldots \ldots .$. from $(2)$
By definition of integration.
$u \cdot w=\int\left[u \cdot v+w \frac{d u}{d x}\right] d x$
$=\int u \cdot v d x+\int w \cdot \frac{d u}{d x} d x$
$\int u \cdot v d x=u \cdot w-\int w \frac{d u}{d x} d x$
$=u \int v d x-\int\left[\frac{d u}{d x} \int v \cdot d x\right] d x$
[next section only required for question 2]
Hence, $\int x e^x d x=x \cdot \int e^x d x-\int\left[\frac{d}{d x} x \cdot \int e^x d x\right] d x$
$=x e^x-\int 1 \times e^x d x$
$=x e^x-e^x+c$