#### Explain Solution R.D.Sharma Class 12 Chapter 18 Indefinite Integrals Exercise 18.19 Question 16 maths Textbook Solution.

Answer: $\frac{14}{9} \log |x-3|+\frac{13}{9} \ln |x+6|+c$

Hint: You know about how to solve integration

Given:$\int \frac{3 x+5}{x^{2}+3 x-18} d x$

Explanation:

We have, $\int \frac{3 x+5}{x^{2}+3 x-18} d x$

\begin{aligned} &\Rightarrow \int \frac{3 x+5}{x^{2}+6 x-3 x-18} d x \\ &\Rightarrow \int \frac{3 x+5}{x(x+6)-3(x+6)} d x \\ &\Rightarrow \int \frac{3 x+5}{(x-3)(x+6)} d x \end{aligned}

Now, Using Partial fraction,

\begin{aligned} &\frac{3 x+5}{(x-3)(x+6)}=\frac{A}{(x+6)}+\frac{B}{(x-3)} \\ &\frac{3 x+5}{(x-3)(x+6)}=\frac{A(x-3)+B(x+6)}{(x-3)(x+6)} \\ &\frac{3 x+5}{(x-3)(x+6)}=\frac{A x-3 A+B x+6 B}{(x-3)(x+6)} \\ &\Rightarrow \frac{3 x+5}{(x-3)(x+6)}=\frac{(A+B) x-3 A+6 B}{(x-3)(x+6)} \end{aligned}

On comparing, we got

$A+B=3$                                                                                            ......(1)

and

\begin{aligned} &-3 A+6 B=5 \\ &(A+B=3) \times-3 \end{aligned}                                                                            ........(2)

We get,

$-3 A-3 B=-9 \ldots .(3)$

Subtract (2) from (3)

\begin{aligned} &(-3 A-3 B)-(-3 A+6 B)=(-9)-(5) \\ &-3 A-3 B+3 A-6 B=-9-5 \\ &-9 B=-14 \\ &B=\frac{14}{9} \end{aligned}

Put value of B in (1)

\begin{aligned} &A+B=3 \\ &A+\frac{14}{9}=3 \\ &A=3-\frac{14}{9} \Rightarrow \frac{27-14}{9} \\ &A=\frac{13}{9} \& B=\frac{14}{9} \end{aligned}

Substituting value of A and B in

\begin{aligned} &\frac{3 x+5}{(x-3)(x+6)}=\frac{A}{(x+6)}+\frac{B}{(x-3)} \\ &\frac{3 x+5}{(x-3)(x+6)}=\frac{13}{9} \frac{1}{(x+6)}+\frac{14}{9} \frac{B}{(x-3)} \end{aligned}

Integrating both sides

$\int \frac{3 x+5}{(x-3)(x+6)} d x=\frac{13}{9} \int \frac{1}{(x+6)} d x+\frac{14}{9} \int \frac{1}{(x-3)} d x$                                                        $\left[\int \frac{1}{x+a} d x=\log |x+a|\right]$

$\int \frac{3 x+5}{(x-3)(x+6)} d x=\frac{13}{9} \log |x+6|+\frac{14}{9} \log |x-3|+c$