What is the integral of 1/(βln^4(β)) ?

The integral of 1/(βln^4(β)) is -1/(3ln^3(β))+C

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integral of 1/(βln^4(β))

\int \frac{1}{β ln ^{4} ( β )} d β

- \frac{1}{3 ln ^{3} ( β )} + C

Solution steps

\int \frac{1}{β ln ^{4} ( β )} d β

Apply u-substitution

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

Apply the Power Rule

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

Substitute back

u = ln (β)

= - \frac{1}{3 ln ^{3} ( β )}

Add a constant to the solution

= - \frac{1}{3 ln ^{3} ( β )} + C

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