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Question

Mathematics Question on integral

If dx(x+2)(x2+1)=alog1+x2+btan1x+15logx+2+C \int\frac{dx}{\left(x+2\right)\left(x^{2} +1\right)} = a log\left|1+x^{2}\right| +b tan^{-1} x +\frac{1}{5}log\left|x+2\right|+C, then

A

a=110,b=25a=\frac{-1}{10} , b=\frac{-2}{5}

B

a=110,b=25a=\frac{1}{10} , b=\frac{-2}{5}

C

a=110,b=25a=\frac{-1}{10} , b=\frac{2}{5}

D

a=110,b=25a=\frac{1}{10} , b=\frac{2}{5}

Answer

a=110,b=25a=\frac{-1}{10} , b=\frac{2}{5}

Explanation

Solution

We have, I=dx(x+2)(x2+1)I = \int\frac{dx}{\left(x+2\right)\left(x^{2}+1\right)} Let 1(x+2)(x2+1)=Ax+2+Bx+Cx2+1 \frac{1}{\left(x+2\right)\left(x^{2}+1\right)} = \frac{A}{x+2} + \frac{Bx+C}{x^{2}+1} 1=A(x2+1)+Bx(x+2)+C(x+2)...(i) \Rightarrow 1= A\left(x^{2} +1\right) + Bx\left(x+2\right) + C\left(x+2\right) \quad...\left(i\right) Put x=0x = 0 in (i)\left(i\right), we get A+2C=1A + 2C = 1 Put x=2x= -2 in (i)\left(i\right), we get A=15A= \frac{1}{5} C=25\Rightarrow C =\frac{2}{5} Put x=1x= 1 in (i)\left(i\right), we get 1=2A+3B+3C 1 = 2A + 3B + 3 C, we get B=15B = \frac{-1}{5} dx(x+2)(x2+1)=15dxx+215xdxx2+1+25dxx2+1 \Rightarrow \int \frac{dx}{\left(x+2\right)\left(x^{2}+1\right)} = \frac{1}{5} \int \frac{dx}{x+2} -\frac{1}{5} \int \frac{xdx}{x^{2}+1} + \frac{2}{5} \int \frac{dx}{x^{2}+1} =15logx+2110logx2+1+25tan1x+C= \frac{1}{5 } log\left|x+2\right| -\frac{1}{10 }log\left|x^{2}+1\right| +\frac{2}{5 }tan^{-1} x + C Hence, a=110a= \frac{-1}{10} and b=25b = \frac{2}{5}