What is the integral of 1/(x^{9/2)} ?

The integral of 1/(x^{9/2)} is -2/(7x^{7/2)}+C

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integral of 1/(x^{9/2)}

\int \frac{1}{x ^{\frac{9}{2}}} d x

- \frac{2}{7 x ^{\frac{7}{2}}} + C

Solution steps

\int \frac{1}{x ^{\frac{9}{2}}} d x

Apply the Power Rule

: - \frac{2}{7 x ^{\frac{7}{2}}}

= - \frac{2}{7 x ^{\frac{7}{2}}}

Add a constant to the solution

= - \frac{2}{7 x ^{\frac{7}{2}}} + C

\frac{6 ( s + 34 )}{s ( s ^{2} + 10 s + 34 )}

f (x) = x^{4} + 5 e^{x}, (0, 5)

f (x) = a x

\int_{- \infty}^{0} x \cdot e^{x} d x

\frac{d x}{d t} = x^{2} e^{- 2 t}

What is the integral of 1/(x^{9/2)} ?
The integral of 1/(x^{9/2)} is -2/(7x^{7/2)}+C