Question

2∫ 1/x(1+x²) dx ,Evaluate it.1

Answer 1

Answer:

The result is:

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

Step-by-step explanation:

Given:

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

Applying partial fraction decomposition method:

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

Comparing the coefficients of $x^2$ on both side of the above equation we get:

$Ax^2+Bx^2=0\\A+B=0\,\,\,\,\,eqn(1)$

Now comparing the coefficients of x from both side of the given equation we get:

$Cx=0\\C=0$

And by comparing the constant terms we get:

$A=1$

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

$A+B=0$

$1+B=0$

$B=-1$

Therefore integrating by substituting the above values:

$\displaystyle\int_1^2 \dfrac{1}{x(1+x^2)}dx=\displaystyle\int_1^2\dfrac{1}{x}dx+\displaystyle\int _1^2\dfrac{-1x+0}{1+x^2}dx\\\displaystyle\int_1^2 \dfrac{1}{x(1+x^2)}dx=\displaystyle\int_1^2 \dfrac{1}{x}dx-\displaystyle\int _1^2\dfrac{1x}{1+x^2}dx$

Let:

$1+x^2=z$

Differentiating with respect to x.

$2x\,dx=dz$

Or,

$x\,dx=\dfrac{dz}{2}$

Substituting for the value of limits:

Lower limit

When $x=1$

$z=1+1\\z=2$

Upper limit:

When $x=2$

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

$\displaystyle\int_1^2 \dfrac{1}{x(1+x^2)}dx=\displaystyle\int _1^2\dfrac{1}{x}dx-\displaystyle\int_2^5 \dfrac{1}{z}\times \dfrac{dz}{2}$

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