What is the integral of (4ln(x)-5)/(3x^5) ?

The integral of (4ln(x)-5)/(3x^5) is (-ln(x)+1)/(3x^4)+C

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integral of (4ln(x)-5)/(3x^5)

\int \frac{4 ln ( x ) - 5}{3 x ^{5}} d x

\frac{- ln ( x ) + 1}{3 x ^{4}} + C

Solution steps

\int \frac{4 ln ( x ) - 5}{3 x ^{5}} d x

Take the constant out:

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

= \frac{1}{3} \cdot \int \frac{4 ln ( x ) - 5}{x ^{5}} d x

Expand

\frac{4 ln ( x ) - 5}{x ^{5}} : \frac{4 ln ( x )}{x ^{5}} - \frac{5}{x ^{5}}

= \frac{1}{3} \cdot \int \frac{4 ln ( x )}{x ^{5}} - \frac{5}{x ^{5}} d x

Apply the Sum Rule:

\int f (x) \pm g (x) d x = \int f (x) d x \pm \int g (x) d x

= \frac{1}{3} (\int \frac{4 ln ( x )}{x ^{5}} d x - \int \frac{5}{x ^{5}} d x)

\int \frac{4 ln ( x )}{x ^{5}} d x = - \frac{ln ( x )}{x ^{4}} - \frac{1}{4 x ^{4}}

\int \frac{5}{x ^{5}} d x = - \frac{5}{4 x ^{4}}

= \frac{1}{3} (- \frac{ln ( x )}{x ^{4}} - \frac{1}{4 x ^{4}} - (- \frac{5}{4 x ^{4}}))

Simplify

\frac{1}{3} (- \frac{ln ( x )}{x ^{4}} - \frac{1}{4 x ^{4}} - (- \frac{5}{4 x ^{4}})) : \frac{- ln ( x ) + 1}{3 x ^{4}}

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

Add a constant to the solution

= \frac{- ln ( x ) + 1}{3 x ^{4}} + C

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What is the integral of (4ln(x)-5)/(3x^5) ?
The integral of (4ln(x)-5)/(3x^5) is (-ln(x)+1)/(3x^4)+C