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  • explain solution RD Sharma class 12 Chapter 18 Indefinite Integrals exercise 18.30 question 22

explain solution RD Sharma class 12 Chapter 18 Indefinite Integrals exercise 18.30 question 22

Answers (1)

Answer:

          \log \left|\frac{(2 \log x+1)^{\square}}{(3 \log x+2)^{1 / 2}}\right|+C

Hint:

            To solve this integration, we use partial fraction method   

Given:

            \int \frac{1}{x\left[6(\log x)^{2}+7 \log x+2\right]} d x

Explanation:

I=\int \frac{1}{x\left[6(\log x)^{2}+7 \log x+2\right]} d x

Let

\begin{aligned} &\log x=y \\ &\frac{1}{x} d x=d y \\ &I=\int \frac{d y}{6 y^{2}+7 y+2} d x \\ &I=\int \frac{d y}{(3 y+2)(2 y+1)} \end{aligned}

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

Comparing the coefficient

2A+3B=0                (1)

A+2B=1                   (2)

Multiply equation (2) by 2 and then

Subtract equation (1) from it

\begin{aligned} &2 A+4 B=2 \quad- \\ &\frac{2 A+3 B=0}{B=2} \end{aligned}

Equation (2)

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

Now

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

As y=\log x

I=\log \left|\frac{(2 \log x+1)^{\square}}{(3 \log x+2)}\right|+C

Posted by

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