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

The integral of 1/((1+x)^4) is -1/(3(1+x)^3)+C

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

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

- \frac{1}{3 ( 1 + x ) ^{3}} + C

Solution steps

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

Apply u-substitution

: \int \frac{1}{u ^{4}} d u

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

Apply the Power Rule

: - \frac{1}{3 u ^{3}}

= - \frac{1}{3 u ^{3}}

Substitute back

u = 1 + x

= - \frac{1}{3 ( 1 + x ) ^{3}}

Add a constant to the solution

= - \frac{1}{3 ( 1 + x ) ^{3}} + C

x \to \frac{π}{2} lim (cot (x))

n = 5 \sum \infty \frac{1}{2 ^{n}}

\frac{d}{d x} (7 x^{2} ln (x))

x^{3} + y^{3} - 3 x^{2} y^{2} + 1

y^{^{''}} + y^{^{'}} - 2 y = 2 t

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