What is the integral of 1/((x^2+a^2)^{1.5)} ?

The integral of 1/((x^2+a^2)^{1.5)} is x/(a^2sqrt(a^2+x^2))+C

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integral of 1/((x^2+a^2)^{1.5)}

\int \frac{1}{( x ^{2} + a ^{2} ) ^{1.5}} d x

\frac{x}{a ^{2} a ^{2} + x ^{2}} + C

Solution steps

\int \frac{1}{( x ^{2} + a ^{2} ) ^{1.5}} d x

(x^{2} + a^{2})^{1.5} = (x^{2} + a^{2}) x^{2} + a^{2}

= \int \frac{1}{( x ^{2} + a ^{2} ) x ^{2} + a ^{2}} d x

Apply Trigonometric Substitution

: \int \frac{1}{a ^{2} sec ( u )} d u

= \int \frac{1}{a ^{2} sec ( u )} d u

Take the constant out:

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

= \frac{1}{a ^{2}} \cdot \int \frac{1}{sec ( u )} d u

Rewrite using trig identities

= \frac{1}{a ^{2}} \cdot \int cos (u) d u

Use the common integral:

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

= \frac{1}{a ^{2}} sin (u)

Substitute back

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

= \frac{1}{a ^{2}} sin (arctan (\frac{1}{a} x))

Simplify

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

= \frac{x}{a ^{2} a ^{2} + x ^{2}}

Add a constant to the solution

= \frac{x}{a ^{2} a ^{2} + x ^{2}} + C

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What is the integral of 1/((x^2+a^2)^{1.5)} ?
The integral of 1/((x^2+a^2)^{1.5)} is x/(a^2sqrt(a^2+x^2))+C