Integrals

Putting sin x = t cos xdx = dt

$\begin{array}{l}{\displaystyle \int \frac{dt}{(1-t)(2-t)}}={\displaystyle \int \frac{(2-t)-(1-t)}{(1-t)(2-t)}}dt.\\ ={\displaystyle \int \frac{dt}{1-t}-{\displaystyle \int \frac{dt}{2-t}}}\\ =\frac{log\left|1-t\right|}{(-1)}-\frac{log\left|2-t\right|}{(-1)}+c\\ =log\left|\frac{2-t}{1-t}\right|+c\\ =log\left|\frac{2-sinx}{1-sinx}\right|+c\end{array}$

$\frac{1}{x\left(x^n+1\right)}=\frac{x^{n-1}}{x^{n-1}\cdot x\left(x^n+1\right)}=\frac{x^{n-1}}{x^n\left(x^n+1\right)}$

Putting xⁿ = t Þ n x^{n- 1}dx = dt

$\begin{array}{l}\therefore {\displaystyle \int \frac{dx}{x\left(x^n+1\right)}}={\displaystyle \int \frac{x^{n-1}dx}{x^n\left(x^n+1\right)}}=\frac{1}{n}{\displaystyle \int \frac{dt}{t(t+1)}}\\ =\frac{1}{n}{\displaystyle \int \frac{(t+1-t)}{t(t+1)}}dt\\ =\frac{1}{n}{\displaystyle \int \left\{\frac{t+1}{t(t+1)}-\frac{t}{t(t+1)}\right\}}dt.\\ =\frac{1}{n}\left\{{\displaystyle \int \frac{dt}{t}}-{\displaystyle \int \frac{dt}{t+1}}\right\}\\ =\frac{1}{n}\{logt-log(t+1)\}+C.\\ =\frac{1}{n}log?\frac{t}{t+1}+\; C\\ =\frac{1}{n}log\left|\frac{x^n}{x^n+1}\right|+C.\end{array}$

Kindly go through the solution

$\begin{array}{l}\Rightarrow A=\raisebox{1ex}{$1$}\!\left/ \!\raisebox{-1ex}{$2$}\right.\\ Andfrom\left(1\right)weget,-\raisebox{1ex}{$-1$}\!\left/ \!\raisebox{-1ex}{$2$}\right.\\ \therefore {\displaystyle \int \frac{dx}{y^2-1}}=\frac{1}{2}{\displaystyle \int \frac{dx}{y-1}}-\frac{1}{2}{\displaystyle \int \frac{dx}{y+1}}\\ =\frac{1}{2}{\displaystyle \int \frac{dx}{x^2-1}}-\frac{1}{2}{\displaystyle \int \frac{dx}{x^2-1}}\left\{y=x^2\right\}\\ =\frac{1}{2}\times \frac{1}{2\cdot 1}log\left|\frac{x-1}{x+1}\right|-\frac{1}{2}ta{n}^{-1}x+\; C\\ =\frac{1}{4}log\left|\frac{x-1}{x+1}\right|-\frac{1}{2}ta{n}^{-1}x+\; C\end{array}$

Kindly go through the solution

Kindly go through the solution

Kindly go through the solution

Kindly go through the solution

Kindly go through the solution

Kindly go through the solution

$\frac{x}{{(x-1)}^2(x-2)}=\frac{A}{(x-1)}+\frac{B}{{(x-1)}^2}+\frac{C}{x+2}$

= A(x- 1)(x + 2) + B(x + 2) + C(x- 1)²

= A(x² + 2x-x- 2) + B(x + 2) + C(x² + 1 - 2x)

= A(x² + x- 2) + B(x + 2) + C(x²- 2x + 1)

Comparing the co-efficient we get, A + C = 0 ¾ (1)

A + B - 2C = 1 ¾ (2)

- 2A + 2B + C = 0 ¾ (3)

EQⁿ (3) - 2 ´ eⁿQ. (2),

- 2A + 2B + C (2A + 2B - 4C) = 0 - 2 ´ 1.

=> - 4A + 5C = - 2