Blog: `int 10/((x - 1)(x^2 + 9)) dx` Evaluate the integral

Integrated `int10/[(x-1)(x^2+9)]dx`

Solve for the variables A, B, and, C using the method of partial fractions.

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

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

`10=Ax^2+9A+Bx^2+Cx-Bx-C`

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

Equate coefficients and solve for A, B, and C.

`0=A+B`

`A=-B`

`0=C-B`

`0=C+A`

`10=9A-C`

`0=A+C`

`10=10A`

`A=1`

`C=-1`

`B=-1`

`int10/[(x-1)(x^2+9)]dx=int[1/(x-1)+(-1x-1)/(x^2+9)]dx`

`=int1/(x-1)dx-intx/(x^2+9)dx-int1/(x^2+9)dx`

The first integral matches the form`int(du)/u=ln|u|+C`

`int1/(x-1)=ln|x-1|+C`

Integrate the second integral using u-substitution.

Let `u=x^2+9`

`(du)/dx=2x`

`dx=(du)/(2x)`

`-intx/(x^2+1)dx=-x/u*(du)/(2x)=-1/2ln|u|+C=-1/2ln|x^2+9|+C`

The third integral matches the form `int(dx)/(x^2+a^2)=1/atan^-1(x/a)+C`

`=-int1/(x^2+9)dx=-1/3tan^-1(x/3)+C`

`=int1/(x-1)dx-intx/(x^2+9)dx-int1/(x^2+9)dx `

`=ln|x-1|-1/2ln(x^2+9)-1/3tan^-1(x/3)+C`

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