What is the integral of (6x^5+x^{1/4}-3)/(-5x^{3/2)} ?

The integral of (6x^5+x^{1/4}-3)/(-5x^{3/2)} is -(4x^5-12x^{1/4}+18)/(15x^{1/2)}+C

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integral of (6x^5+x^{1/4}-3)/(-5x^{3/2)}

\int \frac{6 x ^{5} + x ^{\frac{1}{4}} - 3}{- 5 x ^{\frac{3}{2}}} d x

- \frac{4 x ^{5} - 12 x ^{\frac{1}{4}} + 18}{15 x ^{\frac{1}{2}}} + C

Solution steps

\int \frac{6 x ^{5} + x ^{\frac{1}{4}} - 3}{- 5 x ^{\frac{3}{2}}} d x

Take the constant out:

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

= \frac{1}{- 5} \cdot \int \frac{6 x ^{5} + x ^{\frac{1}{4}} - 3}{x ^{\frac{3}{2}}} d x

Simplify

= (- \frac{1}{5}) \cdot \int \frac{6 x ^{5} + x ^{\frac{1}{4}} - 3}{x ^{\frac{3}{2}}} d x

Expand

\frac{6 x ^{5} + x ^{\frac{1}{4}} - 3}{x ^{\frac{3}{2}}} : 6 x^{\frac{7}{2}} + \frac{1}{x ^{\frac{5}{4}}} - \frac{3}{x ^{\frac{3}{2}}}

= (- \frac{1}{5}) \cdot \int 6 x^{\frac{7}{2}} + \frac{1}{x ^{\frac{5}{4}}} - \frac{3}{x ^{\frac{3}{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

= (- \frac{1}{5}) (\int 6 x^{\frac{7}{2}} d x + \int \frac{1}{x ^{\frac{5}{4}}} d x - \int \frac{3}{x ^{\frac{3}{2}}} d x)

\int 6 x^{\frac{7}{2}} d x = \frac{4}{3} x^{\frac{9}{2}}

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

\int \frac{3}{x ^{\frac{3}{2}}} d x = - \frac{6}{x ^{\frac{1}{2}}}

= (- \frac{1}{5}) (\frac{4}{3} x^{\frac{9}{2}} - \frac{4}{x ^{\frac{1}{4}}} - (- \frac{6}{x ^{\frac{1}{2}}}))

Simplify

(- \frac{1}{5}) (\frac{4}{3} x^{\frac{9}{2}} - \frac{4}{x ^{\frac{1}{4}}} - (- \frac{6}{x ^{\frac{1}{2}}})) : - \frac{4 x ^{5} - 12 x ^{\frac{1}{4}} + 18}{15 x ^{\frac{1}{2}}}

= - \frac{4 x ^{5} - 12 x ^{\frac{1}{4}} + 18}{15 x ^{\frac{1}{2}}}

Add a constant to the solution

= - \frac{4 x ^{5} - 12 x ^{\frac{1}{4}} + 18}{15 x ^{\frac{1}{2}}} + C

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What is the integral of (6x^5+x^{1/4}-3)/(-5x^{3/2)} ?
The integral of (6x^5+x^{1/4}-3)/(-5x^{3/2)} is -(4x^5-12x^{1/4}+18)/(15x^{1/2)}+C