What is the integral of (ln^3(w))/(wsqrt(ln^2(w)-4)) ?

The integral of (ln^3(w))/(wsqrt(ln^2(w)-4)) is 1/3 (ln^2(w)-4)^{3/2}+4sqrt(ln^2(w)-4)+C

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integral of (ln^3(w))/(wsqrt(ln^2(w)-4))

\int \frac{ln ^{3} ( w )}{w ln ^{2} ( w ) - 4} d w

\frac{1}{3} (ln^{2} (w) - 4)^{\frac{3}{2}} + 4 ln^{2} (w) - 4 + C

Solution steps

\int \frac{ln ^{3} ( w )}{w ln ^{2} ( w ) - 4} d w

Apply u-substitution

: \int \frac{u ^{3}}{u ^{2} - 4} d u

= \int \frac{u ^{3}}{u ^{2} - 4} d u

Apply u-substitution

: \int v^{2} + 4 d v

= \int v^{2} + 4 d v

Apply the Sum Rule:

\int f (x) \pm g (x) d x = \int f (x) d x \pm \int g (x) d x

= \int v^{2} d v + \int 4 d v

\int v^{2} d v = \frac{v ^{3}}{3}

\int 4 d v = 4 v

= \frac{v ^{3}}{3} + 4 v

Substitute back

= \frac{( ln ^{2} ( w ) - 4 ) ^{3}}{3} + 4 ln^{2} (w) - 4

Simplify

\frac{( ln ^{2} ( w ) - 4 ) ^{3}}{3} + 4 ln^{2} (w) - 4 : \frac{1}{3} (ln^{2} (w) - 4)^{\frac{3}{2}} + 4 ln^{2} (w) - 4

= \frac{1}{3} (ln^{2} (w) - 4)^{\frac{3}{2}} + 4 ln^{2} (w) - 4

Add a constant to the solution

= \frac{1}{3} (ln^{2} (w) - 4)^{\frac{3}{2}} + 4 ln^{2} (w) - 4 + C

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What is the integral of (ln^3(w))/(wsqrt(ln^2(w)-4)) ?
The integral of (ln^3(w))/(wsqrt(ln^2(w)-4)) is 1/3 (ln^2(w)-4)^{3/2}+4sqrt(ln^2(w)-4)+C