How do you find $int 10/((x-1)(x^2+9)) dx$ using partial fractions?

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Answer 1 · 2015-10-29T21:54:31

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

Since the denominator consists of 1 linear factor and 1 irreducible quadratic factor, we may use the method of partial fractions to rewrite it as :

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

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

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

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

Now comparing terms we get that

$A+B=0$
$C-B=0$
$9A-C=10$

Solving this linear system of equations yields :

$A=1, B=-1, C=-1$

Therefore the original integral may be written and solved as

$int10/((x-1)(x^2+9))dx=int1/(x-1)dx-int(x)/(x^2+9)dx-int1/(x^2+9)dx$

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

How do you find int 10/((x-1)(x^2+9)) dxint 10/((x-1)(x^2+9)) dx using partial fractions?