How do you integrate $\frac{10}{(x - 1) (x^{2} + 9)} d x$ $10/[(x-1)(x^2+9)] dx$ ?

Question

How do you integrate $\frac{10}{(x - 1) (x^{2} + 9)} d x$ $10/[(x-1)(x^2+9)] dx$ ?

1 Answer

Impact of this question

17647 views around the world

You can reuse this answer
Creative Commons License

Answer 1 · 2015-06-17T13:31:02

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

Explanation:

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

So we want: $A x^{2} + 9 A + B x^{2} - B x + C x - C = 10$ $Ax^2+9A +Bx^2-Bx+Cx-C = 10$
Hence:
$A + B = 0$ $A+B =0$
$- B + C = 0$ $-B+C=0$
$9 A - C = 10$ $9A-C=10$

The second equation gives us: $B = C$ $B=C$ , so
$A + C = 0$ $A+C =0$
$9 A - C = 10$ $9A-C=10$ , adding gets us:

$10 A = 10$ $10A=10$ , so $A = 1$ $A=1$ and $B = C = - 1$ $B=C =-1$

And $\frac{10}{(x - 1) (x^{2} + 9)} = \frac{1}{x - 1} + \frac{- x - 1}{x^{2} + 9} = \frac{1}{x - 1} - \frac{x}{x^{2} + 9} - \frac{1}{x^{2} + 9}$ $10/((x-1)(x^2+9)) = 1/(x-1) + (-x-1)/(x^2+9) = 1/(x-1) - x/(x^2+9) - 1/(x^2+9)$ ,

so

$\int \frac{10}{(x - 1) (x^{2} + 9)} d x = \int \frac{1}{x - 1} d x - \int \frac{x}{x^{2} + 9} d x - \int \frac{1}{x^{2} + 9} d x$ $int 10/((x-1)(x^2+9)) dx = int 1/(x-1) dx - int x/(x^2+9) dx - int1/(x^2+9) dx$

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

Science

Math

Humanities

... and beyond

How do you integrate $\frac{10}{(x - 1) (x^{2} + 9)} d x$ $10/[(x-1)(x^2+9)] dx$ ?

1 Answer

Explanation:

Related questions

Impact of this question

Science

Math

Humanities

... and beyond

How do you integrate 10(x−1)(x2+9)dx10/[(x-1)(x^2+9)] dx?

1 Answer

Explanation:

Related questions

Impact of this question

How do you integrate $\frac{10}{(x - 1) (x^{2} + 9)} d x$ $10/[(x-1)(x^2+9)] dx$ ?