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

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

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

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

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

Solution steps

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

Apply Integration By Parts

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

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

Simplify

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

Add a constant to the solution

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

y = sin (2 x), at x = \frac{3 π}{4}

\int_{0}^{π} sin^{4} (x) d x

x^{2} y^{3} d x + x^{3} y^{2} d y = 0

\int \frac{1}{( 1 + x ) ^{3}} d x

x \to \infty lim (2 x^{3})

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