What is the integral of (ln^3(x))/(xsqrt(ln^2(x)-4)) ?

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

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integral of (ln^3(x))/(xsqrt(ln^2(x)-4))

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

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

Solution steps

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

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} ( x ) - 4 ) ^{3}}{3} + 4 ln^{2} (x) - 4

Simplify

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

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

Add a constant to the solution

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

\int \frac{csc ^{2} ( x )}{cot ( x )} d x

\frac{d}{d z} (x e^{z})

\int (x^{2} + 2 x - 10)^{\frac{2}{3}} (5 x + 5) d x

x \to + 0 lim (\frac{f ( x )}{x})

e^{- 2 t} - 3 cos (2 t)

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