#### explain solution RD Sharma class 12 Chapter 18 Indefinite Integrals exercise 18.30 question 15

$\! \! \! \! \! \! \! \! \! \! \! \frac{a\left(a^{2}+b\right)+c}{(a-b)(a-c)} \log |x-a|+\frac{b^{2}(a+1)+c}{(b-a)(b-c)} \log |x-b|+\frac{c(a c+b+1)}{(c-a)(c-b)} \log |x-c|+k$

Hint:

To solve this integration, we use partial fraction method

Given:

$\int \frac{a x^{2}+b x+c}{(x-a)(x-b)(x-c)} d x \\$

Explanation:

\begin{aligned} &I=\int \frac{a x^{2}+b x+c}{(x-a)(x-b)(x-c)} d x \\ &\frac{a x^{2}+b x+c}{(x-a)(x-b)(x-c)}=\frac{A}{x-a}+\frac{B}{x-b}+\frac{C}{x-c} \end{aligned}

\begin{aligned} &\frac{a x^{2}+b x+c}{(x-a)(x-b)(x-c)}=\frac{A(x-b)(x-c)+B(x-a)(x-c)+C(x-a)(x-b)}{(x-a)(x-b)(x-c)} \\ &a x^{2}+b x+c=A(x-b)(x-c)+B(x-a)(x-c)+C(x-a)(x-b) \end{aligned}

\begin{aligned} &\text { For } x=b \\ &a b^{2}+b^{2}+c=0+B(b-a)(b-c)+(0) \\ &\frac{(a+1) b^{2}+c}{(b-a)(b-c)}=B \\ &\text { For } x=a \end{aligned}

\begin{aligned} &a^{3}+b a+c=A(a-b)(a-c)+0+0 \\ &\frac{a^{3}+a b+c}{(a-b)(a-c)}=A \end{aligned}
\begin{aligned} &\text { For } x=c \\ &a c^{2}+b c+c=0+0+C(c-a)(c-b) \\ &\frac{c(a c+b+1)}{(c-a)(c-b)}=C \end{aligned}
\begin{aligned} &\frac{a x^{2}+b x+c}{(x-a)(x-b)(x-c)}=\frac{a\left(a^{2}+b\right)+c}{(a-b)(a-c)} \cdot \frac{1}{x-a}+\frac{b^{2}(a+1)+c}{(b-a)(b-c)} \cdot \frac{1}{x-b}+\frac{c(a c+b+1)}{(c-a)(b-c)} \cdot \frac{1}{x-c} \\ &I=\frac{a\left(a^{2}+b\right)+c}{(a-b)(a-c)} \int \frac{1}{x-a} d x+\frac{b^{2}(a+1)+c}{(b-a)(b-c)} \int \frac{1}{x-b} d x+\frac{c(a c+b+1)}{(c-a)(c-b)} \int \frac{1}{x-c} d x \\ &I=\frac{a\left(a^{2}+b\right)+c}{(a-b)(a-c)} \log |x-a|+\frac{b^{2}(a+1)+c}{(b-a)(b-c)} \log |x-b|+\frac{c(a c+b+1)}{(c-a)(c-b)} \log |x-c|+k \end{aligned}