What is the integral of (3x^2-8x+13)/((x+3)(x-1)^2) ?

The integral of (3x^2-8x+13)/((x+3)(x-1)^2) is 4ln|x+3|-ln|x-1|-2/(x-1)+C

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integral of (3x^2-8x+13)/((x+3)(x-1)^2)

\int \frac{3 x ^{2} - 8 x + 13}{( x + 3 ) ( x - 1 ) ^{2}} d x

4 ln ∣ x + 3 ∣ - ln ∣ x - 1 ∣ - \frac{2}{x - 1} + C

Solution steps

\int \frac{3 x ^{2} - 8 x + 13}{( x + 3 ) ( x - 1 ) ^{2}} d x

Take the partial fraction of

\frac{3 x ^{2} - 8 x + 13}{( x + 3 ) ( x - 1 ) ^{2}} : \frac{4}{x + 3} - \frac{1}{x - 1} + \frac{2}{( x - 1 ) ^{2}}

= \int \frac{4}{x + 3} - \frac{1}{x - 1} + \frac{2}{( 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{4}{x + 3} d x - \int \frac{1}{x - 1} d x + \int \frac{2}{( x - 1 ) ^{2}} d x

\int \frac{4}{x + 3} d x = 4 ln ∣ x + 3 ∣

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

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

= 4 ln ∣ x + 3 ∣ - ln ∣ x - 1 ∣ - \frac{2}{x - 1}

Add a constant to the solution

= 4 ln ∣ x + 3 ∣ - ln ∣ x - 1 ∣ - \frac{2}{x - 1} + C

\frac{\partial}{\partial y} (4 x^{2} + 3 cos (y))

\frac{d}{d x} (x^{2} + \frac{4}{x})

\frac{\partial}{\partial x} (ln (3 x + 3 cy))

cos (5 π x)

\frac{5 - x ^{2}}{5 + x ^{2}}

What is the integral of (3x^2-8x+13)/((x+3)(x-1)^2) ?
The integral of (3x^2-8x+13)/((x+3)(x-1)^2) is 4ln|x+3|-ln|x-1|-2/(x-1)+C