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

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

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

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

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

Solution steps

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

Take the constant out:

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

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

Apply Integration By Parts

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

\int - \frac{1}{3 x ^{4}} d x = \frac{1}{9 x ^{3}}

= 3 (- \frac{ln ( x )}{3 x ^{3}} - \frac{1}{9 x ^{3}})

Simplify

3 (- \frac{ln ( x )}{3 x ^{3}} - \frac{1}{9 x ^{3}}) : - \frac{ln ( x )}{x ^{3}} - \frac{1}{3 x ^{3}}

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

Add a constant to the solution

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

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