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  • explain solution RD Sharma class 12 Chapter 18 Indefinite Integrals exercise 18.30 question 29

explain solution RD Sharma class 12 Chapter 18 Indefinite Integrals exercise 18.30 question 29

Answers (1)

Answer:

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

Hint:

            To solve this integration, we use partial fraction method   

Given:

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

Explanation:

Let

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

\begin{aligned} &x^{2}+1=A(x-2)(x+3)+B(x+3)+C(x-2)^{2} \\ &x^{2}+1=A\left(x^{2}+x-6\right)+B(x+3)+C\left(x^{2}+4-4 x\right) \\ &x^{2}+1=(A+C) x^{2}+(A+B-4 C) x+(-6 A+3 B+4 C) \end{aligned}

Equating the similar terms, we get

A+C=1                        (1)

A+B-4C=0                (2)

-6A+3B+4C=1        (3)

Subtract equation (1) from equation (2) and we get

B-5C=-1                     (4)

Multiply equation (1) by 6 and then adding equation (3)

3B+10C=7                   (5)

Multiply equation (4) by 3 and then subtract it from equation (5)

\! \! \! \! \! \! \! \! 3 B+10 C=7 \\ 3 B-15 C=-3 \\ \overline{25 C=10 }\\\\

\begin{aligned} &C=\frac{10}{25} \\ &C=\frac{2}{5} \end{aligned}

Equation (4)

\begin{aligned} &B-5\left(\frac{2}{5}\right)=-1 \\ &B-2=-1 \\ &B=1 \end{aligned}

Equation (1)

\begin{aligned} &A+\frac{2}{5}=1 \\ &A=1-\frac{2}{5} \\ &A=\frac{3}{5} \end{aligned}

Now

\frac{x^{2}+1}{(x-2)^{2}(x+3)}=\frac{3}{5(x-2)}+\frac{1}{(x-2)^{2}}+\frac{2}{5(x+3)}

Thus

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

Posted by

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