What is the integral of 1/((256+x^2)^{3/2)} ?

The integral of 1/((256+x^2)^{3/2)} is x/(256(256+x^2)^{1/2)}+C

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integral of 1/((256+x^2)^{3/2)}

\int \frac{1}{( 256 + x ^{2} ) ^{\frac{3}{2}}} d x

\frac{x}{256 ( 256 + x ^{2} ) ^{\frac{1}{2}}} + C

Solution steps

\int \frac{1}{( 256 + x ^{2} ) ^{\frac{3}{2}}} d x

Apply Trigonometric Substitution

: \int \frac{1}{256 sec ( u )} d u

= \int \frac{1}{256 sec ( u )} d u

Take the constant out:

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

= \frac{1}{256} \cdot \int \frac{1}{sec ( u )} d u

Rewrite using trig identities

= \frac{1}{256} \cdot \int cos (u) d u

Use the common integral:

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

= \frac{1}{256} sin (u)

Substitute back

u = arctan (\frac{1}{16} x)

= \frac{1}{256} sin (arctan (\frac{1}{16} x))

Simplify

\frac{1}{256} sin (arctan (\frac{1}{16} x)) : \frac{x}{256 ( 256 + x ^{2} ) ^{\frac{1}{2}}}

= \frac{x}{256 ( 256 + x ^{2} ) ^{\frac{1}{2}}}

Add a constant to the solution

= \frac{x}{256 ( 256 + x ^{2} ) ^{\frac{1}{2}}} + C

x \to 0 lim ((3 x^{2} - 2)^{5})

\int e^{x} ln (x) d x

4 x^{3} + 8 x^{2} + 3 x + 2

e^{x^{3}}

\int \frac{1}{( 1 + x )} d x

What is the integral of 1/((256+x^2)^{3/2)} ?
The integral of 1/((256+x^2)^{3/2)} is x/(256(256+x^2)^{1/2)}+C