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  • explain solution RD Sharma class 12 Chapter 18 Indefinite Integrals exercise 18.30 question 63

explain solution RD Sharma class 12 Chapter 18 Indefinite Integrals exercise 18.30 question 63

Answers (1)

Answer:

                x+\frac{2}{3} \tan ^{-1} \frac{x}{\sqrt{3}}-3 \tan ^{-1} \frac{x}{2}

Hint:

            To solve this integration, we use partial fraction method   

Given:

            \int \frac{\left(x^{2}+1\right)\left(x^{2}+2\right)}{\left(x^{2}+3\right)\left(x^{2}+4\right)} d x \\

Explanation:

Let

\begin{aligned} &I=\int \frac{\left(x^{2}+1\right)\left(x^{2}+2\right)}{\left(x^{2}+3\right)\left(x^{2}+4\right)} d x \\ &I=\int \frac{x^{4}+3 x^{2}+2}{x^{4}+7 x^{2}+12} d x \\ &I=\int\left(1-\frac{\left(4 x^{2}+10\right)}{x^{4}+7 x^{2}+12}\right) d x \\ &I=\int 1 d x-\int \frac{4 x^{2}+10}{x^{4}+7 x^{2}+12} d x \\ &I=x-I_{1} \end{aligned}                (1)

Where

\begin{aligned} &I_{1}=\int \frac{4 x^{2}+10}{x^{4}+7 x^{2}+12} d x \\ &\frac{4 x^{2}+10}{x^{4}+7 x^{2}+12}=\frac{4 x^{2}+10}{\left(x^{2}+3\right)\left(x^{2}+4\right)} \end{aligned}              (i)

4 x^{2}+10=(A x+B)\left(x^{2}+4\right)+(C x+D)\left(x^{2}+3\right)

Comparing the coefficient

x^{3} Coefficient

0=A+C                         (2)

x^{2} Coefficient

4=B+D                         (3)

x Coefficient

0=4 A+3 C                     (4)

Constant

10=4 B+3 D                    (5)

\begin{aligned} &\text { Term (2) } A=-C \text { put in (4) }\\ &0=-4 C+3 C\\ &C=0\\ &A=0 \end{aligned}

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

\begin{aligned} &\begin{array}{l} 4 B+4 D=16 \\ 4 B+3 D=10 \\ \hline D=6 \\ B+D=4 \\ B+6=4 \\ B=-2 \end{array} \end{aligned}

(i) Becomes

\begin{aligned} &I_{1}=\int \frac{-2}{x^{2}+3} d x+6 \int \frac{1}{x^{2}+4} d x \\ &I_{1}=-2 \cdot \frac{1}{\sqrt{3}} \tan ^{-1} \frac{x}{\sqrt{3}}+6 \cdot \frac{1}{2} \tan ^{-1} \frac{x}{2} \end{aligned}

\begin{aligned} &I_{1}=\frac{-2}{\sqrt{3}} \tan ^{-1} \frac{x}{\sqrt{3}}+3 \tan ^{-1} \frac{x}{2} \\ &I=x-I_{1} \\ &I=x+\frac{2}{\sqrt{3}} \tan ^{-1} \frac{x}{\sqrt{3}}-3 \tan ^{-1} \frac{x}{2}+C \end{aligned}

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