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

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

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integral of 1/([(1-x^2)(1+x)])

integral

\frac{1}{[ ( 1 - x ^{2} ) ( 1 + x ) ]}

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

Solution steps

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

Simplify

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

Apply u-substitution

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

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

Factor

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

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

Take the constant out:

\int a \cdot f (x) d x = a \cdot \int f (x) d x

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

Take the partial fraction of

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

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

\int \frac{1}{4 u} d u = \frac{1}{4} ln ∣ u ∣

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

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

= - (- \frac{1}{4} ln ∣ u ∣ - (- \frac{1}{2 u}) + \frac{1}{4} ln ∣ u - 2 ∣)

Substitute back

u = 1 + x

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

Simplify

- (- \frac{1}{4} ln ∣ 1 + x ∣ - (- \frac{1}{2 ( 1 + x )}) + \frac{1}{4} ln ∣ 1 + x - 2 ∣) : \frac{1}{4} ln ∣ 1 + x ∣ - \frac{1}{2 ( 1 + x )} - \frac{1}{4} ln ∣ x - 1 ∣

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

Add a constant to the solution

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

\frac{1}{cos ( 2 z ) - 1}

\frac{v + 1}{v ^{2} + 1}

7 x^{6}

(x^{2} + a^{2})^{\frac{1}{2}}

sin (x^{2} + 1)

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