Question

$\int\limits^2_1 1/x(1+x²) dx ,Evaluate it.$

Answer 1

Answer:

The result is:

$log(2)-\dfrac{1}{2}[log(5)-log(2)]$

Step-by-step explanation:

Given equation is:

$\displaystyle\int_1^2\dfrac{1}{x(1+x^2)}dx$

Here we are using partial fraction method:

$\dfrac{1}{x(1+x^2)}=\dfrac{A}{x}+\dfrac{Bx+C}{1+x^2}$

$\dfrac{1}{x(1+x^2)}=\dfrac{A(1+x^2)+(Bx+C)x}{x(1+x^2)}$

$1=A+x^2A+x^2B+xC$

Comparing the coefficients of x^2 we get:

$x^2A+x^2B=0$

$A+B=0\,\,\,\,\,eqn(1)$

Comparing the coefficients of x we get:

$Cx=0$

$C=0$

And by Comparing the coefficients of constant terms we get:

$A=1$

Substituting the value of A in eqn(1) we get:

$B=-1$

Therefore integrating:

$\displaystyle\int _1^2\dfrac{1}{x(1+x^2)}dx=\displaystyle\int _1^2\dfrac{1}{x}dx+\displaystyle\int _1^2\dfrac{-1x+0}{1+x^2}dx$

                         $=\displaystyle\int _1^2\dfrac{1}{x}dx-\displaystyle\int _1^2\dfrac{x}{1+x^2}dx$

Let:

$1+x^2=z\\2x\,dx=dz\\x\,dx=\dfrac{dz}{2}$

Substituting the values limit we get:

Lower limit:

When $x=1$

$z=2$

When $x=2$

$z=(2)^2+1\\z=5$

Hence:

$=\displaystyle\int _1^2\dfrac{1}{x}dx-\displaystyle\int _2^5\dfrac{1}{z}\cdot \dfrac{dz}{2}$

$=[log(x)]_1^2-\dfrac{1}{2}[log{z}]_2^5$

$=log(2)-log(1)-\dfrac{1}{2}[log(5)-log(2)]$

$=log(2)-\dfrac{1}{2}[log(5)-log(2)]$