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  • Need solution for RD Sharma Maths Class 12 Chapter 18 Indefinite Integrals Excercise 18.15 Question 5

Need solution for RD Sharma Maths Class 12 Chapter 18 Indefinite Integrals Excercise 18.15 Question 5

Answers (1)

Answer:

            \frac{1}{2} \tan ^{-1}\left(\frac{x+3}{2}\right)+C

Hint:

            To solve this problem use special integration formula

Given:

            \int \frac{1}{x^{2}+6 x+13} d x

Solution:

Let   I=\int \frac{1}{x^{2}+6 x+13} d x

            \begin{aligned} &=\int \frac{1}{x^{2}+2 \cdot x \cdot 3+3^{2}-3^{2}+13} d x \\ & \end{aligned}

           =\int \frac{1}{(x+3)^{2}-9+13} d x \\

           =\int \frac{1}{(x+3)^{2}+4} d x

Put    x+3=t \Rightarrow d x=d t

Then   I=\int \frac{1}{t^{2}+2^{2}} d t

            =\frac{1}{2} \tan ^{-1}\left(\frac{t}{2}\right)+C                               \left[\because \int \frac{1}{x^{2}+a^{2}} d x=\frac{1}{a} \tan ^{-1}\left(\frac{x}{a}\right)+C\right]

            =\frac{1}{2} \tan ^{-1}\left(\frac{x+3}{2}\right)+C                          \quad[\because t=x+3]

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