How do you integrate $\int \frac{x - 5}{x^{2} (x + 1)}$ $int (x-5) / (x^2(x+1))$ using partial fractions?

Question

How do you integrate $\int \frac{x - 5}{x^{2} (x + 1)}$ $int (x-5) / (x^2(x+1))$ using partial fractions?

1 Answer

Impact of this question

1741 views around the world

You can reuse this answer
Creative Commons License

Answer 1 · 2016-12-21T17:36:13

$6 ln | x | + \frac{5}{x} - 6 ln | x + 1 | + C$ $6ln|x|+5/x-6ln|x+1|+C$

Explanation:

Decompose into partial fractions:
$\frac{x - 5}{x^{2} (x + 1)} \equiv \frac{A x + B}{x^{2}} - \frac{6}{x + 1}$ $(x-5)/(x^2(x+1))-=(Ax+B)/x^2-6/(x+1)$
giving
$x - 5 \equiv (A x + B) (x + 1) - 6 x^{2}$ $x-5-=(Ax+B)(x+1)-6x^2$
Regrouping:
$0 x^{2} + x^{1} - 5 x^{0} \equiv (A - 6) x^{2} + (A + B) x^{1} + B x^{0}$ $0x^2+x^1-5x^0-=(A-6)x^2+(A+B)x^1+Bx^0$
Equating powers of $x$ $x$ :
$A = 6$ $A=6$ , $B = - 5$ $B=-5$
giving the integrand as
$\frac{6 x - 5}{x^{2}} - \frac{6}{x + 1}$ $(6x-5)/x^2-6/(x+1)$
$= \frac{6}{x} - \frac{5}{x^{2}} - \frac{6}{x + 1}$ $=6/x-5/x^2-6/(x+1)$
The $6$ $6$ in the partial fractions comes from the cover-up rule, which avoids getting three simultaneous equations (one for each power of $x$ $x$ in the identity) to solve.

Science

Math

Humanities

... and beyond

How do you integrate $\int \frac{x - 5}{x^{2} (x + 1)}$ $int (x-5) / (x^2(x+1))$ using partial fractions?

1 Answer

Explanation:

Related questions

Impact of this question

Science

Math

Humanities

... and beyond

How do you integrate int (x-5) / (x^2(x+1))∫x−5x2(x+1)int (x-5) / (x^2(x+1)) using partial fractions?

1 Answer

Explanation:

Related questions

Impact of this question

How do you integrate $\int \frac{x - 5}{x^{2} (x + 1)}$ $int (x-5) / (x^2(x+1))$ using partial fractions?