Get Answers to all your Questions

header-bg qa

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

Answers (1)

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

Hint: Using Partial Fraction


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



                          \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


            Multiplying by (x-2)(x+3)


            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}

Posted by


View full answer

Crack CUET with india's "Best Teachers"

  • HD Video Lectures
  • Unlimited Mock Tests
  • Faculty Support