What is the integral of (2x+1)/((x-1)(x+4)^2) ?

The integral of (2x+1)/((x-1)(x+4)^2) is 3/25 ln|x-1|-3/25 ln|x+4|-7/(5(x+4))+C

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integral of (2x+1)/((x-1)(x+4)^2)

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

\frac{3}{25} ln ∣ x - 1 ∣ - \frac{3}{25} ln ∣ x + 4 ∣ - \frac{7}{5 ( x + 4 )} + C

Solution steps

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

Take the partial fraction of

\frac{2 x + 1}{( x - 1 ) ( x + 4 ) ^{2}} : \frac{3}{25 ( x - 1 )} - \frac{3}{25 ( x + 4 )} + \frac{7}{5 ( x + 4 ) ^{2}}

= \int \frac{3}{25 ( x - 1 )} - \frac{3}{25 ( x + 4 )} + \frac{7}{5 ( x + 4 ) ^{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{3}{25 ( x - 1 )} d x - \int \frac{3}{25 ( x + 4 )} d x + \int \frac{7}{5 ( x + 4 ) ^{2}} d x

\int \frac{3}{25 ( x - 1 )} d x = \frac{3}{25} ln ∣ x - 1 ∣

\int \frac{3}{25 ( x + 4 )} d x = \frac{3}{25} ln ∣ x + 4 ∣

\int \frac{7}{5 ( x + 4 ) ^{2}} d x = - \frac{7}{5 ( x + 4 )}

= \frac{3}{25} ln ∣ x - 1 ∣ - \frac{3}{25} ln ∣ x + 4 ∣ - \frac{7}{5 ( x + 4 )}

Add a constant to the solution

= \frac{3}{25} ln ∣ x - 1 ∣ - \frac{3}{25} ln ∣ x + 4 ∣ - \frac{7}{5 ( x + 4 )} + C

\int \frac{1}{x - 1 - x} d x

\int \frac{- 2 y}{x} d x

x \to 5 + lim (\frac{- 3}{x ^{2} - 25})

(1 + x^{2}) y^{^{'}} = x (1 - y)

1 + 4 x - x^{2}, 11 - x - x^{2}, 1 - \frac{2 x}{3}

What is the integral of (2x+1)/((x-1)(x+4)^2) ?
The integral of (2x+1)/((x-1)(x+4)^2) is 3/25 ln|x-1|-3/25 ln|x+4|-7/(5(x+4))+C