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  • explain solution RD Sharma class 12 Chapter 18 Indefinite Integrals exercise 18.30 question 28

explain solution RD Sharma class 12 Chapter 18 Indefinite Integrals exercise 18.30 question 28

Answers (1)

Answer:

            \frac{3}{25} \log \left|\frac{x-3}{x+2}\right|-\frac{7}{5(x-3)}+C

Hint:

            To solve this integration, we use partial fraction method   

Given:

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

Explanation:

Let

\begin{aligned} &I=\int \frac{2 x+1}{(x+2)(x-3)^{2}} d x \\ &\frac{2 x+1}{(x+2)(x-3)^{2}}=\frac{A}{x+2}+\frac{B}{x-3}+\frac{C}{(x-3)^{2}} \\ &2 x+1=A(x-3)^{2}+B(x-3)(x+2)+C(x+2) \\ &2 x+1=x^{2}(A+B)+x(-6 A-B+C)+(9 A-6 B+2 C) \end{aligned}

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

A+B=0                    (1)

\begin{aligned} &A=-B \\ &-6 A-B+C=2 \\ &-6 A+A+C=2 \end{aligned}           [From equation 1]

-5A+C=2                   (2)

\begin{aligned} &9 A-6 B+2 C=1 \\ &9 A+6 A+2 C=1 \\ &15 A+2 C=1 \end{aligned}                   (3)

Multiply the equation (2) by 2 and subtract it from equation (3)

\begin{aligned} &15 A+2 C=1\\ &\frac{-10 A+2 C=4}{25 A=-3}\\ &A=\frac{-3}{25}\\ &B=\frac{3}{25} \end{aligned}                               [From the equation (1)]

Equation (2)

\begin{aligned} &-5\left(\frac{-3}{25}\right)+C=2 \\ &\frac{3}{5}+C=2 \\ &x=2-\frac{3}{5} \\ &C=\frac{7}{5} \end{aligned}

 

\begin{aligned} &\frac{2 x+1}{(x+2)(x-3)^{2}}=\frac{-3}{25(x+2)}+\frac{3}{25(x-3)}+\frac{7}{5(x-3)^{2}} \\ &I=\frac{-3}{25} \int \frac{1}{x+2} d x+\frac{3}{25} \int \frac{1}{x-3}+\frac{7}{5} \int \frac{1}{(x-3)^{2}} \\ &I=\frac{-3}{25} \log |x+2|+\frac{3}{25} \log |x-3|+\frac{7}{5} \cdot \frac{1}{(-1)(x-3)}+C \\ &I=\frac{3}{25} \log \left|\frac{x-3}{x+2}\right|-\frac{7}{5(x-3)}+C \end{aligned}

 

Posted by

infoexpert27

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