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  • Explain Solution R.D.Sharma Class 12 Chapter 18 Indefinite Integrals Exercise Multiple Choice Questions Question 35 Maths Textbook Solution.

Explain Solution R.D.Sharma Class 12 Chapter 18 Indefinite Integrals Exercise Multiple Choice Questions  Question 35 Maths Textbook Solution.

Answers (1)

Answer:

a=\frac{-1}{10}, b=\frac{2}{5}

Given:

\int \frac{1}{(x+2)\left(x^{2}+1\right)} d x=a \log \left|1+x^{2}\right|+b \tan ^{-1} x+\frac{1}{5} \log |x+2|+C

Hint:

Using partial fraction and \int \frac{1}{x} d xand \int \frac{1}{1+x^{2}} d x

Explanation:

Let \mathrm{I}=\int \frac{1}{(x+2)\left(x^{2}+1\right)} d x

Let \frac{1}{(x+2)\left(x^{2}+1\right)}=\frac{A}{x+2}+\frac{B x+C}{x^{2}+1}                        \left [ Eq.(i)) \right ]

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

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

Putting x = -2

\begin{aligned} &1=A(5)+(-2 B+C)(-2+2) \\ &\Rightarrow 1=5 A+0 \\ &\Rightarrow A=\frac{1}{5} \end{aligned}

Putting x = 0

\begin{aligned} &1=A(1)+(B(0)+C)(0+2) \\ &\Rightarrow 1=\frac{1}{5}+2 C \\ &\Rightarrow C=\frac{2}{5} \end{aligned}

Putting x = +1

\begin{aligned} &1=A(1+1)+(B+C)(1+2) \\ &\Rightarrow 1=2\left(\frac{1}{5}\right)+\left(B+\frac{2}{5}\right)(3) \\ &\Rightarrow 1=\frac{2}{5}+3 B+\frac{6}{5} \\ &\Rightarrow 3 B=1-\frac{8}{5} \\ &\Rightarrow 3 B=\frac{-3}{5} \\ &\Rightarrow B=\frac{-1}{5} \end{aligned}

\therefore \text { From }[E q \cdot(i)]

\begin{aligned} &\frac{1}{(x+2)\left(x^{2}+1\right)}=\frac{\frac{1}{5}}{x+2}+\frac{\left(\frac{-1}{5}\right) x+\left(\frac{2}{5}\right)}{x^{2}+1}\\ &\therefore \int \frac{1}{(x+2)\left(x^{2}+1\right)} d x=\frac{1}{5} \int \frac{1}{x+2} d x+\left(\frac{-1}{5}\right) \cdot \frac{1}{2} \int \frac{2 x}{x^{2}+1} d x+\frac{2}{5} \int \frac{1}{x^{2}+1} d x\\ &\therefore I=\frac{1}{5} \log |x+2|-\frac{1}{10} \log \left|1+x^{2}\right|+\frac{2}{5} \tan ^{-1} x+C \end{aligned}\left [ Eq.(ii)) \right ]

Acc. to given

I=a \log \left|1+x^{2}\right|+b \tan ^{-1} x+\frac{1}{5} \log |x+2|                                                                              \left [ Eq.(iii)) \right ]

\begin{aligned} &\therefore \text { From } E q \cdot(i i) \text { and } E q \cdot(\text { iii }) \text { we get }\\ &a=\frac{-1}{10}, b=\frac{2}{5} \end{aligned}

 

 

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