Ncert Solutions Maths class 12th

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

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

=> x = (Ax + B)(x- 1) + C(x² + 1)

= Ax²- Ax + Bx- B + Cx² + C.

Comparing the co-efficients we get,

A + C = 0 ---- (1)

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

- B + C = 0 ---- (3)

Adding (1) and (2) we get,

A + C - A + B = 0 + 1

=> B + C = 1 --- (4)

Adding (4) + (3) we get,

- B + C + B + C = 0 + 1

=> 2C = 1

And C = B from (3)

So, f

$\frac{1-x^2}{x(1-2x)}$ which is not a proper fraction

So, division is missing

$\begin{array}{l}So,\frac{1-x^2}{x(1-2x)}=\frac{1}{2}+\frac{(\raisebox{1ex}{$-x$}\!\left/ \!\raisebox{-1ex}{$2$}\right.+1)}{x(1-2x)}\\ =\frac{1}{2}+\frac{1}{2}\frac{(2-x)}{x(1-2x)}\\ Let\frac{2-x}{x(1-2x)}=\frac{A}{x}+\frac{B}{(1-2x)}\end{array}$

=> 2 -x = A(1 - 2x) + B x.

   So, - 2A + B = - 1

   A = 2.

   So, B = - 1 + 2A = - 1 + 2 ´ 2 = - 1 + 4 = 3

$\begin{array}{l}\therefore {\displaystyle \int \frac{1-x^{2}dx}{x(1-2x)}}={\displaystyle \int \frac{1}{2}}dx+{\displaystyle \int \frac{1}{2}}\frac{(2-x)}{x(1-2x)}dx\\ =\frac{x}{2}+\frac{1}{2}\left\{{\displaystyle \int \frac{A}{x}}dx+{\displaystyle \int \frac{B}{1-2x}}dx\right\}\\ =\frac{x}{2}+\frac{1}{2}\{\left\{\frac{2}{x}dx+{\displaystyle \int \frac{3}{(1-2x)}}dx\right\}\\ =\frac{x}{2}+log\left|x\right|+\frac{3}{2}log\frac{|1-2x|}{(-2)}+c\\ =\frac{x}{2}+log\left|x\right|-\frac{3}{4}log|1-2x|+c\end{array}$