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

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

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

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

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

Solution steps

\int \frac{1}{x 5 + x ^{2}} 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{d}{d x} (5 x sin (4 π) x)

\int 3 x^{2} y^{- 2} + 2 x y^{- 1} d x

\int sin^{4} (\frac{1}{2} x) cos^{4} (\frac{1}{2} x) d x

f (t) = 4 cos (t)

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

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