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

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

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

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

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

Solution steps

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

Apply u-substitution

: \int \frac{1}{u ^{2} ( u - 5 )} d u

= \int \frac{1}{u ^{2} ( u - 5 )} d u

Take the partial fraction of

\frac{1}{u ^{2} ( u - 5 )} : - \frac{1}{25 u} - \frac{1}{5 u ^{2}} + \frac{1}{25 ( u - 5 )}

= \int - \frac{1}{25 u} - \frac{1}{5 u ^{2}} + \frac{1}{25 ( u - 5 )} d u

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}{25 u} d u - \int \frac{1}{5 u ^{2}} d u + \int \frac{1}{25 ( u - 5 )} d u

\int \frac{1}{25 u} d u = \frac{1}{25} ln ∣ u ∣

\int \frac{1}{5 u ^{2}} d u = - \frac{1}{5 u}

\int \frac{1}{25 ( u - 5 )} d u = \frac{1}{25} ln ∣ u - 5 ∣

= - \frac{1}{25} ln ∣ u ∣ - (- \frac{1}{5 u}) + \frac{1}{25} ln ∣ u - 5 ∣

Substitute back

u = x + 4

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

Simplify

- \frac{1}{25} ln ∣ x + 4 ∣ - (- \frac{1}{5 ( x + 4 )}) + \frac{1}{25} ln ∣ x + 4 - 5 ∣ : - \frac{1}{25} ln ∣ x + 4 ∣ + \frac{1}{5 ( x + 4 )} + \frac{1}{25} ln ∣ x - 1 ∣

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

Add a constant to the solution

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

x \to 0 lim (\frac{e ^{x}}{x ^{3}})

\int (3 x - 1)^{5} d x

x \to 0 + lim (\frac{x ^{3}}{ln ( x )})

t \cdot e^{- 3 t}

\frac{\partial}{\partial x} (- 2 x sin (x^{2} + y^{2}))

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