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Need Solution for R.D.Sharma Maths Class 12 Chapter 18 Indefinite Integrals Exercise 18.19 Question 7  Maths Textbook Solution.

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Answer:-\frac{1}{2} \ln \left(3 x^{2}+4 x+2\right)+\frac{3}{\sqrt{2}} \tan ^{-1}\left(\frac{3 x+2}{\sqrt{2}}\right)+c

Hint:Use \int \frac{P n+q}{a x^{2}+b x+c}

Given: \int \frac{1-3 x}{3 x^{2}+4 x+2} d x


I=\int \frac{1-3 x}{3 x^{2}+4 x+2} d n

\int \frac{P n+q}{a x^{2}+b x+c} d x

\text { Numerator }=A\left(\frac{d}{d x}(\text { deno min ator })\right)+B

1-3 x=A\left(\frac{d}{d x}\left(3 x^{2}+4 x+2\right)\right)+B

1-3 x=A(6 x+4)+B

1-3 x=6 A x+4 A+B

On comparing,

6 A=-3 \Rightarrow A=-\frac{3}{6} \Rightarrow A=\frac{-1}{2}

4 A+B=1 \Rightarrow 4\left(\frac{-1}{2}\right)+B=1 \Rightarrow-2+B=1 \Rightarrow B=3

I=\int \frac{\left(-\frac{1}{2}\right)(6 x+4)+3}{3 x^{2}+4 x+2} d x

I=\int-\frac{1}{2} \frac{(6 x+4)}{3 x^{2}+4 x+2} d x+\int \frac{3}{3 x^{2}+4 x+2} d x

=\frac{-1}{2} \int \frac{(6 x+4)}{3 x^{2}+4 x+2} d x+\int \frac{3}{3\left(x^{2}+\frac{4}{3} x+\frac{2}{3}\right)} d x

=\frac{-1}{2} \int \frac{(6 x+4)}{3 x^{2}+4 x+2} d x+\int \frac{1}{x^{2}+\frac{4}{3} x+\frac{4}{9}-\frac{4}{9}+\frac{2}{3}} d x

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

\begin{aligned} &3 x^{2}+4 x+2=t \\ &\frac{d\left(3 x^{2}+4 x+2\right)}{d x}=d t \end{aligned}

=-\frac{1}{2} \int \frac{d t}{t}+\frac{1}{\frac{\sqrt{2}}{3}}\left[\frac{\tan ^{-1} x+\frac{2}{3}}{\frac{\sqrt{2}}{3}}\right]\left[\int \frac{1}{x^{2}+a^{2}}=\frac{1}{a} \tan ^{-1} \frac{x}{2}+c\right]

=-\frac{1}{2} \ln \left(3 x^{2}+4 x+2\right)+\frac{3}{\sqrt{2}} \tan ^{-1}\left(\frac{3 x+2}{\sqrt{2}}\right)+c


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