#### Need Solution for R.D.Sharma Maths Class 12 Chapter 18 Indefinite Integrals Exercise 18.19 Question 9 Maths Textbook Solution.

Answer:$\frac{a}{4} \ln \left|x^{4}+c^{2}\right|+\frac{b}{2 c} \tan ^{-1} \frac{x^{2}}{c}+f$

Hint : Find $I_{1}\: and \: I_{2}$

Given: $\int \frac{a x^{3}+b x}{x^{4}+c^{2}} d x$

Solution: $\int \frac{a x^{3}+b x}{x^{4}+c^{2}} d x$

\begin{aligned} I &=\int \frac{a x^{3}}{x^{4}+c^{2}} d x+\int \frac{b x}{x^{4}+c^{2}} d x \\ I_{1} &=\int \frac{a x^{3}}{x^{4}+c^{2}} d x \end{aligned}

Let,$x^{4}+c^{2}=t$

\begin{aligned} &4 x^{3} d x=d t \\ &x^{3} d x=\frac{1}{4} d t \end{aligned}

$\Rightarrow I_{1}=\frac{a}{4} \int \frac{d t}{t}=\frac{a}{4} \ln t+C_{1}$

\begin{aligned} &I_{2}=\int \frac{b x}{x^{4}+c^{2}} d x \\ &\text { Let } x^{2}=v \\ &2 x d x=d v \end{aligned}

$I_{2}=\frac{b}{2} \int \frac{d v}{v^{2}+c^{2}}$

$=\frac{b}{2} \tan ^{-1}\left(\frac{v}{c}\right) \times \frac{1}{c}+c_{2} \quad\left[\int \frac{1}{x^{2}+a^{2}}=\frac{1}{a} \tan ^{-1} \frac{x}{2}+c\right]$

$I_{2}=\frac{b}{2 c} \tan ^{-1}\left(\frac{x^{2}}{c}\right)+c_{2}$

\begin{aligned} &I=I_{1}+I_{2} \\ &I=\frac{a}{4} \ln \left|x^{4}+c^{2}\right|+c_{1}+\frac{b}{2 c} \tan ^{-1}\left(\frac{x^{2}}{c}\right)+c_{2} \\ &c_{1}+c_{2}=f \\ &I=\frac{a}{4} \ln \left|x^{4}+c^{2}\right|+\frac{b}{2 c} \tan ^{-1}\left(\frac{x^{2}}{c}\right)+f \end{aligned}