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

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

Answers (1)

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

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