What is the integral of 1/(xsqrt(x^2+5)) ?

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

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integral of 1/(xsqrt(x^2+5))

\int \frac{1}{x x ^{2} + 5} d x

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

Solution steps

\int \frac{1}{x x ^{2} + 5} 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{1}{5} x )}{2}

Add a constant to the solution

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

\frac{\partial}{\partial y} (\frac{x ^{2}}{2})

f (x) = x 3 - x

\int 5 x ln (3 - x) d x

y = (3 x + 1)^{2}, (0, 1)

f (x) = x^{5}

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