Maths Solve the Following Question.(2 Marks)

Let $I=\int \frac{1}{x \cdot \log x \log (\log x)} d x$
Put log (log x) = t
Differentiating w.r.t. x, we get
$\frac{1}{\log x} \frac{.1}{x} d x=d t$
$I=\int \frac{1}{t} d t=\log |t|+c$
$I=\log |\log (\log x)|+c$

$\int u d v=u v-\int v d u$
Choosing u = logx and dv = xdx
$d u=\frac{1}{x} d x$
$v=\frac{x^2}{2}$
$\therefore \int x \log x d x=\log x \frac{x^2}{2}-\int \frac{x^2}{2} \frac{1}{x} d x$
$=\frac{x^2}{2} \log x-\frac{1}{2} \int x d x$
$=\frac{x^2}{2} \log x-\frac{1}{2} \frac{x^2}{2}+C$
$=\frac{x^2}{2} \log x-\frac{x^2}{4}+C$

Consider the integral, $\int \frac{d x}{x^2+4 x+8}$
$I =\int \frac{1}{x^2+4 x+8} dx$
$=\int \frac{1}{x^2+4 x+4+4} dx$
$=\int \frac{1}{(x+2)^2+2^2} d x \quad\{$ Use Intergral Formula :}
$\int \frac{1}{x^2+a^2} d x=\frac{1}{a} \tan ^{-1}\left(\frac{x}{a}\right)$
$=\frac{1}{2} \tan ^{-1} \frac{x+2}{2}+C$

$\begin{aligned} & I=\int e^x\left[\frac{\cos x}{\sin ^2 x}-\frac{\sin x}{\sin ^2 x}\right] d x \\ & =\int e^x\left[\begin{array}{cc}\cot x \cdot \cos e c x & -\cos e c x \\ f \prime(x) & f(x)\end{array}\right] \\ & \because \int e^x[f(x)+f \prime(x)] d x=e^x f(x)+C \\ & \therefore I=-e^x \cdot \cos e c x+C\end{aligned}$

$\begin{aligned} & \text { Let } I=\int\left(\frac{\sin \sqrt{x}}{\sqrt{x}}\right) d x \\ & \text { Let } \sqrt{x}=t \\ & \frac{1}{\sqrt{x}}=\frac{d t}{d x} \\ & \frac{1}{\sqrt{x}} d x=2 d t \\ & \therefore I=2 \int \sin t d t \\ & =-2 \cos t+C \\ & =-2 \cos (\sqrt{x})+C\end{aligned}$

$\int \frac{\sin x}{\sqrt{36-\cos ^2 x}} d x$
Substitute, cosx = t
∴ - sin x dx = dt
∴ sin x dx = - dt
The integral becomes
$\begin{aligned} & \int \frac{-d t}{\sqrt{36-t^2}} \\ & =-\int \frac{d t}{\sqrt{6^2-t^2}} \\ & =-\sin ^{-1}\left(\frac{t}{6}\right)+C \\ & =-\sin ^{-1} \cdot\left(\frac{\cos x}{6}\right)+c\end{aligned}$

$\begin{aligned} & \int e^x\left[\frac{\sqrt{1-x^2} \sin ^{-1} x+1}{\sqrt{1-x^2}}\right] d x \\ & =\int e^x\left[\sin ^{-1} x+\frac{1}{\sqrt{1-x^2}}\right] d x\end{aligned}$
$\begin{aligned} & \text { We know that } \int e^x[f(x)+f \prime(x)] d x=e^x \cdot f(x)+c \\ & =e^x \cdot \sin ^{-1} x+c\end{aligned}$

$I=\int \sec ^{n-1} x \sec x \tan x d x$
Let secx=t
$\begin{aligned} & \therefore \sec x \tan x d x=d t \\ & I=\int t^{n-1} d t \\ & =\frac{t^n}{n}+c \\ & =\frac{\sec ^n x}{n}+C\end{aligned}$

$\begin{aligned} & \text { Let } I =\int \frac{x+1}{(x+2)(x+3)} d x \\ & \frac{x+1}{(x+2)(x+3)}=\frac{A}{x+2}+\frac{B}{x+3} \\ & x+1=A(x+3)+B(x+2) \ldots \ldots . . .(1)\end{aligned}$
∴ Putting x = -2 in equation (i) we get
-1 = A
∴ A = -1
∴ Putting x = -3 in equation (i) we get
-2 = -B
∴ B = 2
∴(x+1)/((x+2)(x+3))=1/(x+2)+2/(x+3)
$\begin{aligned} & \therefore I=\int\left[-\frac{1}{x+2}+\frac{2}{x+3}\right] d x \\ & \therefore I=-\log |x+2|+2 \log |x+3|+c\end{aligned}$