How do you use partial fractions to find the integral $\int \frac{6 x^{2} + 1}{x^{2} {(x - 1)}^{3}} d x$ $int (6x^2+1)/(x^2(x-1)^3)dx$ ?

Question

How do you use partial fractions to find the integral $\int \frac{6 x^{2} + 1}{x^{2} {(x - 1)}^{3}} d x$ $int (6x^2+1)/(x^2(x-1)^3)dx$ ?

1 Answer

Impact of this question

1490 views around the world

You can reuse this answer
Creative Commons License

Answer 1 · 2017-10-12T17:00:50

$- 3 ln x + \frac{1}{x} - 3 ln (x - 1) + \frac{2}{x - 1} - \frac{7}{2} \cdot {(x - 1)}^{- 2} + C$ $-3lnx+1/x-3ln(x-1)+2/(x-1)-7/2*(x-1)^(-2)+C$

Explanation:

I decomposed integrand into basic fractions,

$\frac{6 x^{2} + 1}{x^{2} \cdot {(x - 1)}^{2}}$ $(6x^2+1)/[x^2*(x-1)^2]$

= $- \frac{3}{x} - \frac{1}{x^{2}} + \frac{3}{x - 1} - \frac{2}{{(x - 1)}^{2}} + \frac{7}{{(x - 1)}^{3}}$ $-3/x-1/x^2+3/(x-1)-2/(x-1)^2+7/(x-1)^3$

Hence,

$\int \frac{(6 x^{2} + 1) d x}{x^{2} \cdot {(x - 1)}^{3}}$ $int ((6x^2+1)dx)/[x^2*(x-1)^3]$

= $- \int \frac{3 d x}{x}$ $-int (3dx)/x$ - $\int \frac{d x}{x^{2}}$ $int dx/x^2$ - $\int \frac{3 d x}{x - 1}$ $int (3dx)/(x-1)$ - $\int \frac{2 d x}{{(x - 1)}^{2}}$ $int (2dx)/(x-1)^2$ + $\int \frac{7 d x}{{(x - 1)}^{3}}$ $int (7dx)/(x-1)^3$

= $- 3 ln x + \frac{1}{x} - 3 ln (x - 1) + \frac{2}{x - 1} - \frac{7}{2} \cdot {(x - 1)}^{- 2} + C$ $-3lnx+1/x-3ln(x-1)+2/(x-1)-7/2*(x-1)^(-2)+C$

Science

Math

Humanities

... and beyond

How do you use partial fractions to find the integral $\int \frac{6 x^{2} + 1}{x^{2} {(x - 1)}^{3}} d x$ $int (6x^2+1)/(x^2(x-1)^3)dx$ ?

1 Answer

Explanation:

Related questions

Impact of this question

Science

Math

Humanities

... and beyond

How do you use partial fractions to find the integral ∫6x2+1x2(x−1)3dxint (6x^2+1)/(x^2(x-1)^3)dx?

1 Answer

Explanation:

Related questions

Impact of this question

How do you use partial fractions to find the integral $\int \frac{6 x^{2} + 1}{x^{2} {(x - 1)}^{3}} d x$ $int (6x^2+1)/(x^2(x-1)^3)dx$ ?