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  • Explain Solution R.D.Sharma Class 12 Chapter 18 Indefinite Integrals Exercise Revision Exercise Question 124 Maths Textbook Solution.

Explain Solution R.D.Sharma Class 12 Chapter 18 Indefinite Integrals Exercise Revision Exercise  Question 124 Maths Textbook Solution.

Answers (1)

Answer: \frac{1}{2} \log \left|\frac{1+x}{\sqrt{1+x^{2}}}\right|+\frac{1}{2} \tan ^{-1} x+C

Given:\int \frac{1}{1+x+x^{2}+x^{3}} d x

Hint: using partial function

Explanation: let I=\int \frac{1}{1+x+x^{2}+x^{2}} d x

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

Multiplying by  (1+x)\left(1+x^{2}\right)

1=A\left(1+x^{2}\right)+(B x+C)(1+x)

\begin{aligned} &\text { put } x=-1 \\ &1=A(1+1)+(B(-1)+C)(1-1) \\ &1=A(2)+0=>A=\frac{1}{2} \end{aligned}

\begin{aligned} &\text { put } x=0 \\ &1=A(1+0)+(B(0)+C)(1+0) \\ &1=A+C=>1-\frac{1}{2}=C \Rightarrow \frac{1}{2} \end{aligned}

\begin{aligned} &\text { putting } x=1 \\ &1=A(1+1)+(B+C)(1+1) \\ &1=2 A+2 B+2 C \\ &1=2\left(\frac{1}{2}\right)+2 B+2\left(\frac{1}{2}\right) \\ &1=2 B+1=>2 B=1-2=>B=-\frac{1}{2} \end{aligned}

\begin{aligned} &\frac{1}{(1+x)\left(1+x^{2}\right)}=\frac{1}{2(1+x)}+\frac{1}{2}\left(\frac{-x+1}{\left(1+x^{2}\right)}\right) \\ &\int \frac{1}{(1+x)\left(1+x^{2}\right)} d x=\frac{1}{2} \int \frac{1}{1+x} d x-\frac{1}{2} \int \frac{x-1}{1+x^{2}} d x \\ &=\frac{1}{2} \int \frac{1}{1+x^{2}} d x-\frac{1}{2^{2}} \int \frac{2 x}{1+x^{2}} d x+\frac{1}{2} \int \frac{1}{1+x^{2}} d x \\ &=\frac{1}{2} \log |1+x|-\frac{1}{4} \log \left|1+x^{2}\right|+\frac{1}{2} \tan ^{-1} x+C \\ &=\frac{1}{2}\left[\log |1+x|-\log \left|\sqrt{1+x^{2}}\right|\right]+\frac{1}{2} \tan ^{-1} x+C \\ &=\frac{1}{2} \log \left|\frac{1+x}{\sqrt{1+x^{2}}}\right|+\frac{1}{2} \tan ^{-1} x+C \end{aligned}

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