What is the integral of 1/(xsqrt(ax^2+\sqrt{a))} ?

The integral of 1/(xsqrt(ax^2+\sqrt{a))} is 1/(\sqrt[4]{a)}ln|tan((arctan(\sqrt[4]{a}x))/2)|+C

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integral of 1/(xsqrt(ax^2+\sqrt{a))}

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

\frac{1}{4 a} ln tan (\frac{arctan ( 4 a x )}{2}) + C

Solution steps

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

Apply Trigonometric Substitution

: \int \frac{sec ( u )}{4 a tan ( u )} d u

= \int \frac{sec ( u )}{4 a tan ( u )} d u

Take the constant out:

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

= \frac{1}{4 a} \cdot \int \frac{sec ( u )}{tan ( u )} d u

Apply u-substitution

: \int \frac{1}{v} d v

= \frac{1}{4 a} \cdot \int \frac{1}{v} d v

Use the common integral:

\int \frac{1}{v} d v = ln (∣ v ∣)

= \frac{1}{4 a} ln ∣ v ∣

Substitute back

= \frac{1}{4 a} ln tan \frac{arctan ( \frac{a}{4 a} x )}{2}

Simplify

\frac{1}{4 a} ln tan \frac{arctan ( \frac{a}{4 a} x )}{2} : \frac{1}{4 a} ln tan (\frac{arctan ( 4 a x )}{2})

= \frac{1}{4 a} ln tan (\frac{arctan ( 4 a x )}{2})

Add a constant to the solution

= \frac{1}{4 a} ln tan (\frac{arctan ( 4 a x )}{2}) + C

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What is the integral of 1/(xsqrt(ax^2+\sqrt{a))} ?
The integral of 1/(xsqrt(ax^2+\sqrt{a))} is 1/(\sqrt[4]{a)}ln|tan((arctan(\sqrt[4]{a}x))/2)|+C