What is the integral of 1/(xsqrt(9x^2+25)) ?

The integral of 1/(xsqrt(9x^2+25)) is 1/5 ln|tan((arctan(3/5 x))/2)|+C

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integral of 1/(xsqrt(9x^2+25))

\int \frac{1}{x 9 x ^{2} + 25} d x

\frac{1}{5} ln tan (\frac{arctan ( \frac{3}{5} x )}{2}) + C

Solution steps

\int \frac{1}{x 9 x ^{2} + 25} d x

Apply Trigonometric Substitution

: \int \frac{sec ( u )}{5 tan ( u )} d u

= \int \frac{sec ( u )}{5 tan ( u )} d u

Take the constant out:

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

= \frac{1}{5} \cdot \int \frac{sec ( u )}{tan ( u )} d u

Apply u-substitution

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

= \frac{1}{5} \cdot \int \frac{1}{v} d v

Use the common integral:

\int \frac{1}{v} d v = ln (∣ v ∣)

= \frac{1}{5} ln ∣ v ∣

Substitute back

= \frac{1}{5} ln tan (\frac{arctan ( \frac{3}{5} x )}{2})

Add a constant to the solution

= \frac{1}{5} ln tan (\frac{arctan ( \frac{3}{5} x )}{2}) + C

3 y^{^{''}} - 8 y^{^{'}} - 3 y = 0

\int \frac{5}{36 - 25 x ^{2}} d x

x \to + 0 lim (\frac{sin ( + ( x ) )}{x})

\int sin^{4} (x) cos^{6} (x) d x

y^{^{'}} = - \frac{4 t}{y}, y (0) = 2

What is the integral of 1/(xsqrt(9x^2+25)) ?
The integral of 1/(xsqrt(9x^2+25)) is 1/5 ln|tan((arctan(3/5 x))/2)|+C