What is the integral of 1/(xsqrt(64+9x^2)) ?

The integral of 1/(xsqrt(64+9x^2)) is 1/8 ln|tan((arctan(3/8 x))/2)|+C

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integral of 1/(xsqrt(64+9x^2))

\int \frac{1}{x 64 + 9 x ^{2}} d x

\frac{1}{8} ln tan (\frac{arctan ( \frac{3}{8} x )}{2}) + C

Solution steps

\int \frac{1}{x 64 + 9 x ^{2}} d x

Apply Trigonometric Substitution

: \int \frac{sec ( u )}{8 tan ( u )} d u

= \int \frac{sec ( u )}{8 tan ( u )} d u

Take the constant out:

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

= \frac{1}{8} \cdot \int \frac{sec ( u )}{tan ( u )} d u

Apply u-substitution

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

= \frac{1}{8} \cdot \int \frac{1}{v} d v

Use the common integral:

\int \frac{1}{v} d v = ln (∣ v ∣)

= \frac{1}{8} ln ∣ v ∣

Substitute back

= \frac{1}{8} ln tan (\frac{arctan ( \frac{3}{8} x )}{2})

Add a constant to the solution

= \frac{1}{8} ln tan (\frac{arctan ( \frac{3}{8} x )}{2}) + C

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What is the integral of 1/(xsqrt(64+9x^2)) ?
The integral of 1/(xsqrt(64+9x^2)) is 1/8 ln|tan((arctan(3/8 x))/2)|+C