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

The integral of 1/((x+3)^2(x-1)) is -1/16 ln|x+3|+1/(4(x+3))+1/16 ln|x-1|+C

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

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

- \frac{1}{16} ln ∣ x + 3 ∣ + \frac{1}{4 ( x + 3 )} + \frac{1}{16} ln ∣ x - 1 ∣ + C

Solution steps

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

Apply u-substitution

: \int \frac{1}{u ^{2} ( u - 4 )} d u

= \int \frac{1}{u ^{2} ( u - 4 )} d u

Take the partial fraction of

\frac{1}{u ^{2} ( u - 4 )} : - \frac{1}{16 u} - \frac{1}{4 u ^{2}} + \frac{1}{16 ( u - 4 )}

= \int - \frac{1}{16 u} - \frac{1}{4 u ^{2}} + \frac{1}{16 ( u - 4 )} 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}{16 u} d u - \int \frac{1}{4 u ^{2}} d u + \int \frac{1}{16 ( u - 4 )} d u

\int \frac{1}{16 u} d u = \frac{1}{16} ln ∣ u ∣

\int \frac{1}{4 u ^{2}} d u = - \frac{1}{4 u}

\int \frac{1}{16 ( u - 4 )} d u = \frac{1}{16} ln ∣ u - 4 ∣

= - \frac{1}{16} ln ∣ u ∣ - (- \frac{1}{4 u}) + \frac{1}{16} ln ∣ u - 4 ∣

Substitute back

u = x + 3

= - \frac{1}{16} ln ∣ x + 3 ∣ - (- \frac{1}{4 ( x + 3 )}) + \frac{1}{16} ln ∣ x + 3 - 4 ∣

Simplify

- \frac{1}{16} ln ∣ x + 3 ∣ - (- \frac{1}{4 ( x + 3 )}) + \frac{1}{16} ln ∣ x + 3 - 4 ∣ : - \frac{1}{16} ln ∣ x + 3 ∣ + \frac{1}{4 ( x + 3 )} + \frac{1}{16} ln ∣ x - 1 ∣

= - \frac{1}{16} ln ∣ x + 3 ∣ + \frac{1}{4 ( x + 3 )} + \frac{1}{16} ln ∣ x - 1 ∣

Add a constant to the solution

= - \frac{1}{16} ln ∣ x + 3 ∣ + \frac{1}{4 ( x + 3 )} + \frac{1}{16} ln ∣ x - 1 ∣ + C

y = 9 x - 2 x^{2}, (- 2, - 26)

\frac{d}{d x} (\frac{6 x}{6 + x})

\frac{\partial}{\partial x} (\frac{( x + y )}{y})

y = - 0.0959 x^{2} + 106.83 x + 71149

\int 2 e^{t x} x d x

What is the integral of 1/((x+3)^2(x-1)) ?
The integral of 1/((x+3)^2(x-1)) is -1/16 ln|x+3|+1/(4(x+3))+1/16 ln|x-1|+C