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  • explain solution RD Sharma class 12 Chapter 18 Indefinite Integrals exercise 18.30 question 9

explain solution RD Sharma class 12 Chapter 18 Indefinite Integrals exercise 18.30 question 9

Answers (1)

Answer:

            \frac{-1}{10} \log |x-1|+\frac{5}{2} \log |x+1|-\frac{12}{5} \log |2 x+3|+C

Hint:

            To solve this integration, we use partial fraction method   

Given:

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

Explanation:

Let

I=\int \frac{2 x-3}{(x-1)(x+1)(2 x+3)}…(applying the formula a^{2}-b^{2})

\frac{2 x-3}{\left(x^{2}-1\right)(2 x+3)}=\frac{2 x-3}{(x-1)(x+1)(2 x+3)}

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

\begin{aligned} &\text { At } x=1 \\ &2 \times 1-3=A(2)(5)+B(0)+C(0) \\ &-1=10 \mathrm{~A} \\ &A=\frac{-1}{10} \end{aligned}

 

\begin{aligned} &\text { At } x=-1 \\ &2 \times-1-3=A(0)+B(-2)(1)+0 \\ &-5=-2 B \\ &B=\frac{5}{2} \end{aligned}

\begin{aligned} &\text { At } x=\frac{-3}{2} \\ &2 \times\left(\frac{-3}{2}\right)-3=A(0)+B(0)+C\left(\frac{-3}{2}+1\right)\left(\frac{-3}{2}-1\right) \\ &-6=C\left(\frac{5}{4}\right) \\ &C=\frac{-24}{5} \end{aligned}
\frac{2 x-3}{(x-1)(x+1)(2 x+3)}=\frac{-1}{10(x-1)}+\frac{5}{2(x+1)}-\frac{24}{5(2 x+3)}

 

 

\begin{aligned} &I=\frac{-1}{10} \int \frac{1}{x-1} x+\frac{5}{2} \int \frac{1}{x+1} d x-\frac{24}{5} \int \frac{1}{2 x+3} d x \\ &I=\frac{-1}{10} \log |x-1|+\frac{5}{2} \log |x+1|-\frac{12}{5} \log |2 x+3|+C \\ &I=\frac{-1}{10} \log |x-1|+\frac{5}{2} \log |x+1|-\frac{12}{5} \log |2 x+3|+C \end{aligned}
 

 

 

Posted by

infoexpert27

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