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  • explain solution RD Sharma class 12 Chapter 18 Indefinite Integrals exercise 18.30 question 14

explain solution RD Sharma class 12 Chapter 18 Indefinite Integrals exercise 18.30 question 14

Answers (1)

Answer:

            \frac{3}{5} \log |x+2|+\frac{1}{5} \log \left|x^{2}+1\right|+\frac{1}{5} \tan ^{-1} x+C

Hint:

            To solve this integration, we use partial fraction method   

Given:

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

Explanation:

Let

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

Comparing the coefficient

A+B=1           (1)

2B+C=1        (2)

A+2C=1         (3)

Subtract equation (3) from equation (1), we get

\begin{aligned} &A+B=1 \quad- \\ &\frac{A+2 C=1}{B-2 C=0} \\ &B=2 C \end{aligned}

Equation (2)

\begin{aligned} &2(2 C)+C=1 \\ &4 C+C=1 \\ &5 C=1 \\ &C=\frac{1}{5} \\ &B=\frac{2}{5} \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}+x+1}{\left(x^{2}+1\right)(x+2)}=\frac{3}{5(x+2)}+\frac{\frac{2}{5} x+\frac{1}{5}}{\left(x^{2}+1\right)}

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

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

Posted by

infoexpert27

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