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

The integral of (ln(x))/(x^a) is 1/(-a+1)x^{-a+1}ln(x)-(x^{-a+1})/((-a+1)^2)+C

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integral of (ln(x))/(x^a)

\int \frac{ln ( x )}{x ^{a}} d x

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

Solution steps

\int \frac{ln ( x )}{x ^{a}} d x

Apply Integration By Parts

: \frac{1}{- a + 1} x^{- a + 1} ln (x) - \int \frac{1}{x ^{a} ( - a + 1 )} d x

= \frac{1}{- a + 1} x^{- a + 1} ln (x) - \int \frac{1}{x ^{a} ( - a + 1 )} d x

\int \frac{1}{x ^{a} ( - a + 1 )} d x = \frac{x ^{- a + 1}}{( - a + 1 ) ^{2}}

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

Add a constant to the solution

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

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