What is the integral of 1/((64+x^2)^{3/2)} ?

The integral of 1/((64+x^2)^{3/2)} is x/(64(64+x^2)^{1/2)}+C

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integral of 1/((64+x^2)^{3/2)}

\int \frac{1}{( 64 + x ^{2} ) ^{\frac{3}{2}}} d x

\frac{x}{64 ( 64 + x ^{2} ) ^{\frac{1}{2}}} + C

Solution steps

\int \frac{1}{( 64 + x ^{2} ) ^{\frac{3}{2}}} d x

Apply Trigonometric Substitution

: \int \frac{1}{64 sec ( u )} d u

= \int \frac{1}{64 sec ( u )} d u

Take the constant out:

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

= \frac{1}{64} \cdot \int \frac{1}{sec ( u )} d u

Rewrite using trig identities

= \frac{1}{64} \cdot \int cos (u) d u

Use the common integral:

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

= \frac{1}{64} sin (u)

Substitute back

u = arctan (\frac{1}{8} x)

= \frac{1}{64} sin (arctan (\frac{1}{8} x))

Simplify

\frac{1}{64} sin (arctan (\frac{1}{8} x)) : \frac{x}{64 ( 64 + x ^{2} ) ^{\frac{1}{2}}}

= \frac{x}{64 ( 64 + x ^{2} ) ^{\frac{1}{2}}}

Add a constant to the solution

= \frac{x}{64 ( 64 + x ^{2} ) ^{\frac{1}{2}}} + C

\int_{0}^{1} (\frac{x}{1 + x ^{2}}) d x

x y^{^{'}} + y = 55 x

\int \frac{5 sin ( 2 x )}{sin ( x )} d x

y = 1 0^{n x}

\int \frac{e ^{- 3 x}}{x} d x

What is the integral of 1/((64+x^2)^{3/2)} ?
The integral of 1/((64+x^2)^{3/2)} is x/(64(64+x^2)^{1/2)}+C