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

The integral of ((ln(x))^2)/(4x^3) is (-2ln^2(x)-2ln(x)-1)/(16x^2)+C

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

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

\frac{- 2 ln ^{2} ( x ) - 2 ln ( x ) - 1}{16 x ^{2}} + C

Solution steps

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

Take the constant out:

\int a \cdot f (x) d x = a \cdot \int f (x) d x

= \frac{1}{4} \cdot \int \frac{( ln ( x ) ) ^{2}}{x ^{3}} d x

Apply Integration By Parts

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

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

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

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

Simplify

\frac{1}{4} (- \frac{ln ^{2} ( x )}{2 x ^{2}} - (\frac{ln ( x )}{2 x ^{2}} + \frac{1}{4 x ^{2}})) : \frac{- 2 ln ^{2} ( x ) - 2 ln ( x ) - 1}{16 x ^{2}}

= \frac{- 2 ln ^{2} ( x ) - 2 ln ( x ) - 1}{16 x ^{2}}

Add a constant to the solution

= \frac{- 2 ln ^{2} ( x ) - 2 ln ( x ) - 1}{16 x ^{2}} + C

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What is the integral of ((ln(x))^2)/(4x^3) ?
The integral of ((ln(x))^2)/(4x^3) is (-2ln^2(x)-2ln(x)-1)/(16x^2)+C