What is the integral of (x^m-x^n)/(sqrt(x)) ?

The integral of (x^m-x^n)/(sqrt(x)) is (2x^{(2m+1)/2})/(2m+1)-(2x^{(2n+1)/2})/(2n+1)+C

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integral of (x^m-x^n)/(sqrt(x))

\int \frac{x ^{m} - x ^{n}}{x} d x

\frac{2 x ^{\frac{2 m + 1}{2}}}{2 m + 1} - \frac{2 x ^{\frac{2 n + 1}{2}}}{2 n + 1} + C

Solution steps

\int \frac{x ^{m} - x ^{n}}{x} d x

Expand

\frac{x ^{m} - x ^{n}}{x} : x^{\frac{2 m - 1}{2}} - x^{\frac{2 n - 1}{2}}

= \int x^{\frac{2 m - 1}{2}} - x^{\frac{2 n - 1}{2}} 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

= \int x^{\frac{2 m - 1}{2}} d x - \int x^{\frac{2 n - 1}{2}} d x

\int x^{\frac{2 m - 1}{2}} d x = \frac{2 x ^{\frac{2 m + 1}{2}}}{2 m + 1}

\int x^{\frac{2 n - 1}{2}} d x = \frac{2 x ^{\frac{2 n + 1}{2}}}{2 n + 1}

= \frac{2 x ^{\frac{2 m + 1}{2}}}{2 m + 1} - \frac{2 x ^{\frac{2 n + 1}{2}}}{2 n + 1}

Add a constant to the solution

= \frac{2 x ^{\frac{2 m + 1}{2}}}{2 m + 1} - \frac{2 x ^{\frac{2 n + 1}{2}}}{2 n + 1} + C

What is the integral of (x^m-x^n)/(sqrt(x)) ?
The integral of (x^m-x^n)/(sqrt(x)) is (2x^{(2m+1)/2})/(2m+1)-(2x^{(2n+1)/2})/(2n+1)+C