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

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

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

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

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

Solution steps

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

Take the constant out:

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

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

Apply Integration By Parts

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

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

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

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

Simplify

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

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

Add a constant to the solution

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

\int 2 x + 6 d x

\frac{1}{s ^{4} ( s + 3 )}

\frac{\partial}{\partial z} (\frac{z}{x ^{2} + y ^{2} + z ^{2}})

\frac{d y}{d x} + 15 x^{2} y = 45 x^{2}

\frac{d}{d x} (\frac{1}{1 - 5 x})

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