What is the integral of (2x^2+x+1)/((x+3)(x-1)^2) ?

The integral of (2x^2+x+1)/((x+3)(x-1)^2) is ln|x+3|+ln|x-1|-1/(x-1)+C

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integral of (2x^2+x+1)/((x+3)(x-1)^2)

\int \frac{2 x ^{2} + x + 1}{( x + 3 ) ( x - 1 ) ^{2}} d x

ln ∣ x + 3 ∣ + ln ∣ x - 1 ∣ - \frac{1}{x - 1} + C

Solution steps

\int \frac{2 x ^{2} + x + 1}{( x + 3 ) ( x - 1 ) ^{2}} d x

Take the partial fraction of

\frac{2 x ^{2} + x + 1}{( x + 3 ) ( x - 1 ) ^{2}} : \frac{1}{x + 3} + \frac{1}{x - 1} + \frac{1}{( x - 1 ) ^{2}}

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

Apply the Sum Rule:

\int f (x) \pm g (x) d x = \int f (x) d x \pm \int g (x) d x

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

\int \frac{1}{x + 3} d x = ln ∣ x + 3 ∣

\int \frac{1}{x - 1} d x = ln ∣ x - 1 ∣

\int \frac{1}{( x - 1 ) ^{2}} d x = - \frac{1}{x - 1}

= ln ∣ x + 3 ∣ + ln ∣ x - 1 ∣ - \frac{1}{x - 1}

Add a constant to the solution

= ln ∣ x + 3 ∣ + ln ∣ x - 1 ∣ - \frac{1}{x - 1} + C

n = 1 \sum \infty 1 6^{n} x^{n} n!

\int x x^{2} + 7 d x

\int \frac{1}{2000 ( 2000 - x )} d x

d er i v a t i v e f (x) = x^{a} (1 - x)^{b}

t an g e n t f (x) = 9 - 4 x + 6 x^{2}, at x = 8

What is the integral of (2x^2+x+1)/((x+3)(x-1)^2) ?
The integral of (2x^2+x+1)/((x+3)(x-1)^2) is ln|x+3|+ln|x-1|-1/(x-1)+C