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Need Solution for R.D.Sharma Maths Class 12 Chapter 18 Indefinite Integrals  Exercise 18.19 Question 4  Maths Textbook Solution.

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I=\log \left|x^{2}+6 x+13\right|-\frac{9}{2} \tan ^{-1}\left(\frac{x+3}{2}\right)+c

Hint: Using Formula \int \frac{d x}{x^{2}-a^{2}}=\frac{1}{a} \tan ^{-1}\left(\frac{x}{a}\right)+c \& \int \frac{d t}{t}=\log t+c

Given:  \int \frac{2 x-3}{x^{2}+6 x+13} d x

Solution:          \frac{d}{d x}\left(x^{2}+6 x+13\right)=2 x+6

Let,           2 x-3=A(2 x+6)+B

                2 x-3=2 A x+6 A+B

On comparing,

2 x=2 A x=>A=1

6 A+B=-3=>B=-9

I=\int \frac{2 x+6}{x^{2}+6 x+13} d x-9 \int \frac{d x}{x^{2}+6 x+13}

\text { Let, } x^{2}+6 x+13=t

               (2 x+6) d x=d t

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

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

I=\int \frac{d t}{t}-9 \int \frac{d x}{(x+3)^{2}-4}

I=\int \frac{d t}{t}-9 \int \frac{d x}{(x+3)^{2}-2^{2}}                                                    \left[\int \frac{d x}{x^{2}-a^{2}}=\frac{1}{a} \tan ^{-1}\left(\frac{x}{a}\right)+c \& \int \frac{d t}{t}=\log t+c\right]

I=\log |t|-9 \times \frac{1}{2} \tan ^{-1}\left(\frac{x+3}{2}\right)+c

I=\log \left|x^{2}+6 x+13\right|-\frac{9}{2} \tan ^{-1}\left(\frac{x+3}{2}\right)+c

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