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

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Answer 1 · 2015-10-29T21:54:31

$ln | x - 1 | - \frac{1}{2} ln ∣ ∣ x^{2} + 9 ∣ ∣ + \frac{1}{3} {tan}^{- 1} (\frac{x}{3}) + C$ $ln|x-1|-1/2ln|x^2+9|+1/3tan^(-1)(x/3)+C$

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 :

$\frac{10}{(x - 1) (x^{2} + 9)} = \frac{A}{x - 1} + \frac{B x + C}{x^{2} + 9}$ $10/((x-1)(x^2+9))=A/(x-1)+(Bx+C)/(x^2+9)$

$= \frac{A (x^{2} + 9) + (B x + C) (x - 1)}{(x - 1) (x^{2} + 9)}$ $=(A(x^2+9)+(Bx+C)(x-1))/((x-1)(x^2+9))$

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

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

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)) dx∫10(x−1)(x2+9)dxint 10/((x-1)(x^2+9)) dx using partial fractions?