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

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

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

integral

\frac{1}{( 1 + t ) ^{4}}

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

Solution steps

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

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 + t

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

Add a constant to the solution

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

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

cos^{3} (x) sin^{2} (x)

x^{5} (2 - x^{3})^{\frac{1}{2}}

9 x^{2} sin (4 x)

\frac{1}{( e ^{t} - 2 )}

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