What is the integral of 1/((x+4)^{7/2)} ?

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

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integral of 1/((x+4)^{7/2)}

\int \frac{1}{( x + 4 ) ^{\frac{7}{2}}} d x

- \frac{2}{5 ( x + 4 ) ^{\frac{5}{2}}} + C

Solution steps

\int \frac{1}{( x + 4 ) ^{\frac{7}{2}}} d x

Apply u-substitution

: \int \frac{1}{u ^{\frac{7}{2}}} d u

= \int \frac{1}{u ^{\frac{7}{2}}} d u

Apply the Power Rule

: - \frac{2}{5 u ^{\frac{5}{2}}}

= - \frac{2}{5 u ^{\frac{5}{2}}}

Substitute back

u = x + 4

= - \frac{2}{5 ( x + 4 ) ^{\frac{5}{2}}}

Add a constant to the solution

= - \frac{2}{5 ( x + 4 ) ^{\frac{5}{2}}} + C

y^{^{'}} + 2 x y = 1

(- 2)^{^{'}}

\frac{\partial}{\partial x} (e^{t})

x \to 1 lim (x^{3} - 2 x + 2)

\int \frac{2}{4 x + 7 x} d x

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