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

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

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

\int \frac{( ln ( x ) ) ^{2}}{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 ( 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{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^{^{'}} = - 2 x e^{2 x}, y (0) = 1

\frac{1}{s ^{1} ( s + 1 ) ^{3}}

f (x) = x^{2} - 4, - 3, 1

\int x x d x

\frac{d}{d t} (a cos (2 t) + b sin (2 t))

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