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  • explain solution RD Sharma class 12 Chapter 18 Indefinite Integrals exercise 18.30 question 43

explain solution RD Sharma class 12 Chapter 18 Indefinite Integrals exercise 18.30 question 43

Answers (1)

Answer:

                \frac{1}{2} \log \left|\frac{x+1}{x-1}\right|-\frac{4}{x-1}+C

Hint:

            To solve this integration, we use partial fraction method   

Given:

            \int \frac{3 x+5}{x^{3}-x^{2}-x+1} d x \\

Explanation:

Let

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

\begin{aligned} &I=\int \frac{3 x+5}{(x+1)(x-1)^{2}} d x \\ &\frac{3 x+5}{(x+1)(x-1)^{2}}=\frac{A}{x-1}+\frac{B}{(x-1)^{2}}+\frac{C}{x+1} \\ &3 x+5=A(x-1)(x+1)+B(x+1)+C(x-1)^{2} \end{aligned}           (1)

Put x= 1in equation (1)

\begin{aligned} &3+5=A(0)+B(2)+C(0) \\ &8=2 B \\ &B=4 \end{aligned}

Put x= -1 in equation (1)

\begin{aligned} &-3+5=A(0)+B(0)+C(4) \\ &2=4 C \\ &C=\frac{1}{2} \end{aligned}

Put x=0in equation (1)

\begin{aligned} &5=A(-1)(1)+B+C \\ &5=-A+B+C \\ &5=-A+4+\frac{1}{2} \\ &1=-A+\frac{1}{2} \end{aligned}

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

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

 

Posted by

infoexpert27

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