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  • explain solution RD Sharma class 12 Chapter 18 Indefinite Integrals exercise 18.30 question 27

explain solution RD Sharma class 12 Chapter 18 Indefinite Integrals exercise 18.30 question 27

Answers (1)

Answer:

            \frac{11}{4} \log \left|\frac{x+1}{x+3}\right|+\frac{5}{2(x+1)}+C

Hint:

            To solve this integration, we use partial fraction method   

Given:

            \int \frac{3 x-2}{(x+1)^{2}(x+3)} d x    

Explanation:

Let

\begin{aligned} &I=\int \frac{3 x-2}{(x+1)^{2}(x+3)} d x \\ &\frac{3 x-2}{(x+1)^{2}(x+3)}=\frac{A}{(x+1)}+\frac{B}{(x+1)^{2}}+\frac{C}{x+3} \\ &\frac{3 x-2}{(x+1)^{2}(x+3)}=\frac{A(x+1)(x+3)+B(x+3)+C(x+1)^{2}}{(x+1)^{2}(x+3)} \end{aligned}

\begin{aligned} &3 x-2=A\left(x^{2}+4 x+3\right)+B(x+3)+C\left(x^{2}+1+2 x\right) \\ &3 x-2=(A+C) x^{2}+x(4 A+B+2 C)+(3 A+3 B+C) \end{aligned}

Comparing the coefficient of x^{2},xand constant term

A+C=0

A=-C                               (1)

4A+B+2C=3             (2)

-4C+B+2C=3          from equation (1)

B-2C=3                        (3)

\begin{aligned} &3 A+3 B+C=-2 \\ &-3 C+3 B+C=-2 \end{aligned}       from equation (1)

3B-2C=-2                  (4)

Subtract equation (3) from equation (4)

\begin{aligned} &3 B-2 C=-2- \\ &B-2 C=3 \\ &\overline{ 2 B=-5} \\ &B=\frac{-5}{2} \end{aligned}

Equation (3)

\begin{aligned} &\frac{-5}{2}-2 C=3 \\ &\frac{-5}{2}-3=2 C \\ &\frac{-11}{2}=2 C \\ &C=\frac{-11}{4} \\ &A=\frac{11}{4} \end{aligned}
\frac{3 x-2}{(x+1)^{2}(x+3)}=\frac{11}{4(x+1)}-\frac{5}{2(x+1)^{2}}-\frac{11}{4(x+3)}

Thus

\begin{aligned} &I=\frac{11}{4} \int \frac{1}{x+1} d x-\frac{5}{2} \int \frac{1}{(x+1)^{2}} d x-\frac{11}{4} \int \frac{1}{x+3} d x \\ &I=\frac{11}{4} \log |x+1|-\frac{5}{2}\left(\frac{1}{-1(x+1)}\right)-\frac{11}{4} \log |x+3|+C \\ &I=\frac{11}{4} \log \left|\frac{x+1}{x+3}\right|+\frac{5}{2(x+1)}+C \end{aligned}

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