What is the integral of 1/(xsqrt(4+x^2)) ?

The integral of 1/(xsqrt(4+x^2)) is 1/2 ln|tan((arctan(1/2 x))/2)|+C

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integral of 1/(xsqrt(4+x^2))

\int \frac{1}{x 4 + x ^{2}} d x

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

Solution steps

\int \frac{1}{x 4 + x ^{2}} d x

Apply Trigonometric Substitution

: \int \frac{sec ( u )}{2 tan ( u )} d u

= \int \frac{sec ( u )}{2 tan ( u )} d u

Take the constant out:

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

= \frac{1}{2} \cdot \int \frac{sec ( u )}{tan ( u )} d u

Apply u-substitution

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

= \frac{1}{2} \cdot \int \frac{1}{v} d v

Use the common integral:

\int \frac{1}{v} d v = ln (∣ v ∣)

= \frac{1}{2} ln ∣ v ∣

Substitute back

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

Add a constant to the solution

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

y^{^{'}} = k y

y^{^{''}} - y^{^{'}} - 2 y = 4 x^{2}, y (0) = - 1, y^{^{'}} (0) = 1

x \to 16 lim (4 x)

\int_{0}^{5} π (\frac{20 - 4 x}{5})^{2} d x

\int 3 u d u

What is the integral of 1/(xsqrt(4+x^2)) ?
The integral of 1/(xsqrt(4+x^2)) is 1/2 ln|tan((arctan(1/2 x))/2)|+C