What is the integral of 1/((625+x^2)^{3/2)} ?

The integral of 1/((625+x^2)^{3/2)} is x/(625(625+x^2)^{1/2)}+C

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integral of 1/((625+x^2)^{3/2)}

\int \frac{1}{( 625 + x ^{2} ) ^{\frac{3}{2}}} d x

\frac{x}{625 ( 625 + x ^{2} ) ^{\frac{1}{2}}} + C

Solution steps

\int \frac{1}{( 625 + x ^{2} ) ^{\frac{3}{2}}} d x

Apply Trigonometric Substitution

: \int \frac{1}{625 sec ( u )} d u

= \int \frac{1}{625 sec ( u )} d u

Take the constant out:

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

= \frac{1}{625} \cdot \int \frac{1}{sec ( u )} d u

Rewrite using trig identities

= \frac{1}{625} \cdot \int cos (u) d u

Use the common integral:

\int cos (u) d u = sin (u)

= \frac{1}{625} sin (u)

Substitute back

u = arctan (\frac{1}{25} x)

= \frac{1}{625} sin (arctan (\frac{1}{25} x))

Simplify

\frac{1}{625} sin (arctan (\frac{1}{25} x)) : \frac{x}{625 ( 625 + x ^{2} ) ^{\frac{1}{2}}}

= \frac{x}{625 ( 625 + x ^{2} ) ^{\frac{1}{2}}}

Add a constant to the solution

= \frac{x}{625 ( 625 + x ^{2} ) ^{\frac{1}{2}}} + C

\int 2 + 2 cos (θ) d θ

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\int 6 e^{7 x} d x

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y^{^{''}} - 8 y^{^{'}} + 16 y = 0, y (0) = 1, y^{^{'}} (0) = 2

What is the integral of 1/((625+x^2)^{3/2)} ?
The integral of 1/((625+x^2)^{3/2)} is x/(625(625+x^2)^{1/2)}+C