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

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

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

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

ln (\frac{1}{2} x^{2} + 4 x + x + 2) + C

Solution steps

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

Complete the square

x^{2} + 4 x : (x + 2)^{2} - 4

= \int \frac{1}{( x + 2 ) ^{2} - 4} d x

Apply u-substitution

: \int \frac{1}{u ^{2} - 4} d u

= \int \frac{1}{u ^{2} - 4} d u

Apply Trigonometric Substitution

: \int sec (v) d v

= \int sec (v) d v

Use the common integral:

\int sec (v) d v = ln ∣ tan (v) + sec (v) ∣

= ln ∣ tan (v) + sec (v) ∣

Substitute back

= ln tan (arcsec (\frac{1}{2} (x + 2))) + sec (arcsec (\frac{1}{2} (x + 2)))

Simplify

ln tan (arcsec (\frac{1}{2} (x + 2))) + sec (arcsec (\frac{1}{2} (x + 2))) : ln (\frac{1}{2} x^{2} + 4 x + x + 2)

= ln (\frac{1}{2} x^{2} + 4 x + x + 2)

Add a constant to the solution

= ln (\frac{1}{2} x^{2} + 4 x + x + 2) + C

y^{^{''}} - y^{^{'}} - 2 y = - 6 t + 6 t^{2}

\int \frac{1}{2 x ^{3} + x} d x

\frac{\partial}{\partial x} ((0.5 x^{2} + 0.5 y^{2})^{\frac{1}{2}})

\int_{0}^{\frac{π}{2}} cos^{5} (1) d x

y^{^{''}} - 4 y^{^{'}} + 4 y = t^{- 6} e^{2 t}

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