Question

How do I find the integral `int10/((x-1)(x^2+9))dx` ?

Answer 1

`=5/4ln(x-1)-5/8ln(x^2+9)+5/12tan^-1(x/3)+c`, where `c` is a constant

Explanation :

This type of question usually solve by using Partial Fractions,

`10/((x-1)(x^2+9))`, it can be written as

`10/((x-1)(x^2+9))=A/(x-1)+(Bx+C)/(x^2+9)`

multiplying both sides by `(x-1)(x^2+9)`, we get

`10=A(x^2+9)+(Bx+C)(x-1)`

`10=(A+B)x^2+(C-B)x+(9A-C)`

comparing coefficients of `x^2`, `x` and constant both side,

`A+B=0` `=>` `A=-B` .................`(i)`

`C-B=0` `=>` `C=-B` .................`(ii)`

`9A-C=10` `=>` .................`(iii)`

substituting `A` & `C`from `(i)` and `(ii)` in `(iii)`

`9(-B)-(-B)=10`

`B=-10/8=-5/4`

Therefore, `A=C=5/4`

Hence, we get

`10/((x-1)(x^2+9))=5/(4(x-1))+((-5/4)x+(5/4))/(x^2+9)`

integrating both sides with respect to `x`,

`int10/((x-1)(x^2+9))dx=int5/(4(x-1))dx+int((-5/4)x+(5/4))/(x^2+9)dx`

`=5/4int1/((x-1))dx-5/4int(x)/(x^2+9)dx+5/4int1/(x^2+9)dx`

Second integral is solved by substitution, as the numerator is the differentiation of denominator

`=5/4ln(x-1)-5/4*1/2ln(x^2+9)+5/4*1/3tan^-1(x/3)`

`=5/4ln(x-1)-5/8ln(x^2+9)+5/12tan^-1(x/3)+c`, where `c` is a constant