Need Solution for R.D.Sharma Maths Class 12 Chapter 18 Indefinite Integrals Exercise 18.20 Question 2 Maths Textbook Solution.

Given: $x+\log \left|\frac{x-2}{x+3}\right|+c$

Hint: Using Partial Fraction

Explanation:

$I=\int \frac{x^{2}+x-1}{x^{2}+x-6} d x$

$\frac{x^{2}+x-1}{x^{2}+x-6}=\frac{x^{2}+x-6+5}{x^{2}+x-6}=1+\frac{5}{x^{2}+x-6}$

$=1+\frac{5}{(x-2)(x+3)}$

$\left[\begin{array}{l} x^{2}+x-6=x^{2}+3 x-2 x-6 \\ =x(x+3)-2(x+3) \\ =(x-2)(x+3) \end{array}\right]$

$\because \int \frac{x^{2}+x-1}{x^{2}+x-6} d x=\int 1 d x+5 \int \frac{1}{(x-2)(x+3)} d x$

$\because I=x+5 I_{1}$ .......................(1)

Where $I_{1}=\int \frac{1}{(x-2)(x+3)} d x$

$\frac{1}{(x-2)(x+3)}=\frac{A}{(x-2)}+\frac{B}{(x+3)}$

Multiplying by $(x-2)(x+3)$

$1=A(x+3)+B(x-2)$

Put x=3

$1=A(0)+B(-5) \Rightarrow B=\frac{-1}{5}$

Put x=2

$1=A(5)+B(0) \Rightarrow A=\frac{1}{5}$

$\therefore \frac{1}{(x-2)(x+3)}=\frac{\frac{1}{5}}{(x-2)}+\frac{-\frac{1}{5}}{(x+3)}$

$\therefore \int \frac{1}{(x-2)(x+3)} d x=\frac{1}{5} \int \frac{1}{x-2} d x+\left(\frac{-1}{5}\right) \int \frac{1}{x+3} d x$

\begin{aligned} &=\frac{1}{5} \log |x-2|-\frac{1}{5} \log |x+3|+c \\ &=\frac{1}{5} \log \left|\frac{x-2}{x+3}\right|+c \end{aligned}

Put in (1)

\begin{aligned} &I=x+5\left(\frac{1}{5} \log \left|\frac{x-2}{x+3}\right|\right)+c \\ &I=x+\log \left|\frac{x-2}{x+3}\right|+c \end{aligned}