What is the integral of 7x^{2/5}+6x^{-4/5} ?

The integral of 7x^{2/5}+6x^{-4/5} is 5x^{7/5}+30x^{1/5}+C

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integral of 7x^{2/5}+6x^{-4/5}

\int 7 x^{\frac{2}{5}} + 6 x^{- \frac{4}{5}} d x

5 x^{\frac{7}{5}} + 30 x^{\frac{1}{5}} + C

Solution steps

\int 7 x^{\frac{2}{5}} + 6 x^{- \frac{4}{5}} d x

6 x^{- \frac{4}{5}} = \frac{6}{x ^{\frac{4}{5}}}

= \int 7 x^{\frac{2}{5}} + \frac{6}{x ^{\frac{4}{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

= \int 7 x^{\frac{2}{5}} d x + \int \frac{6}{x ^{\frac{4}{5}}} d x

\int 7 x^{\frac{2}{5}} d x = 5 x^{\frac{7}{5}}

\int \frac{6}{x ^{\frac{4}{5}}} d x = 30 x^{\frac{1}{5}}

= 5 x^{\frac{7}{5}} + 30 x^{\frac{1}{5}}

Add a constant to the solution

= 5 x^{\frac{7}{5}} + 30 x^{\frac{1}{5}} + C

\frac{d}{d x} (- \frac{1}{x ^{5}} - \frac{1}{x ^{3}} + \frac{1}{x})

\int \frac{2 t - 1}{t ^{2} - 2 t + 1} d t

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

\int \frac{1}{u 1 + u ^{2}} d u

\frac{d}{d x} (7 sin (337 x + \frac{7 π}{18}))

What is the integral of 7x^{2/5}+6x^{-4/5} ?
The integral of 7x^{2/5}+6x^{-4/5} is 5x^{7/5}+30x^{1/5}+C