Question

3
∫ dx/x²(x+1) ,Evaluate it.
1

Answer 1

Answer:

The result is $\dfrac{2}{3}+log(2)-log(3)$.

Step-by-step explanation:

Given:

$\displaystyle\int_1^3\dfrac{dx}{x^2(x+1)}$

Here we are applying formula of partial fraction:

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

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

$Ax^2+Cx^2=0\\A+C\,\,\,\,\,eqn(1)$

Comparing the coefficient of $x$ from both side of the equation we get:

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

And by comparing the constant terms we get:

$B=1$

Hence substituting the value of $B$ in eqn(2) we get:

$A=-1$

And again substituting the value of $A$ in eqn(1) we get:

$C=1$

Now integrating by substituting the above values:

$\displaystyle\int_1^3 \dfrac{1}{x^2(x+1)}=\displaystyle\int_1^3 \dfrac{-1}{x}dx+\displaystyle\int_1^3 \dfrac{1}{x^2}dx+\displaystyle\int_1^3 \dfrac{dx}{x+1}$

Let:

$x+1=z$

Differentiating with respect to x.

$dx=dz$

Now substituting for the value of limits:

Lower limit:

When $x=1$

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

Upper limit:

When $x=3$

$z=3+1\\z=4$

Therefore evaluating it:

$=\displaystyle\int_1^3 \dfrac{-1}{x}dx+\displaystyle\int_1^3 \dfrac{1}{x^2}dx+\displaystyle\int_2^4\dfrac{dz}{z}\\=-[log(x)]_1^3+(\dfrac{-1}{x})_1^3+[log(z)]_2^4\\=-[log(3)-log(1)]-(\dfrac{1}{3}-1)+[log(4)-log(2)]\\=-[log(3)-0]-(\dfrac{-2}{3})+[2log(2)-log(2)]\\=\dfrac{2}{3}-log(3)+log(2)$