#### Explain Solution R.D.Sharma Class 12 Chapter 18 Indefinite Integrals Exercise 18.20 Question 10 maths Textbook Solution.

Answer: $x-3 \log \left|x^{2}+6 x+12\right|+2 \sqrt{3} \tan ^{-1}\left(\frac{x+3}{\sqrt{3}}\right)+c$

Given: $\int \frac{x^{2}}{x^{2}+6 x+12} d x$

Hint: Using $\int \frac{1}{x} d x \text { and } \int \frac{1}{1+x^{2}} d x$

Explanation: Let

$I=\int \frac{x^{2}}{x^{2}+6 x+12} d x$

\begin{aligned} &=\int\left(1-\left(\frac{6 x+12}{x^{2}+6 x+12}\right)\right) d x \\ &=\int 1 d x-6 \int \frac{x+2}{x^{2}+6 x+12} d x \\ &=\int 1 d x-\frac{6}{2} \int \frac{2 x+4+2-2}{x^{2}+6 x+12} d x \end{aligned}

\begin{aligned} &=\int 1 d x-3 \int \frac{2 x+6}{x^{2}+6 x+12} d x+6 \int \frac{1}{x^{2}+6 x+(3)^{2}-(3)^{2}+12} d x \\ &=\int 1 d x-3 \int \frac{2 x+6}{x^{2}+6 x+12}+6 \int \frac{1}{(x+3)^{2}+(\sqrt{3})^{2}} d x \end{aligned}

\begin{aligned} &=x-3 \log \left|x^{2}+6 x+12\right|+6 \times \frac{1}{\sqrt{3}} \tan ^{-1}\left(\frac{x+3}{\sqrt{3}}\right)+c \\ &=x-3 \log \left|x^{2}+6 x+12\right|+2 \sqrt{3} \tan ^{-1}\left(\frac{x+3}{\sqrt{3}}\right)+c \end{aligned}