What is the integral of 1/(sqrt(24x+6x^2)) ?

The integral of 1/(sqrt(24x+6x^2)) is 1/(sqrt(6))ln((|sqrt(x^2+4x)+x+2|)/2)+C

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integral of 1/(sqrt(24x+6x^2))

\int \frac{1}{24 x + 6 x ^{2}} d x

\frac{1}{6} ln (\frac{x ^{2} + 4 x + x + 2}{2}) + C

Solution steps

\int \frac{1}{24 x + 6 x ^{2}} d x

Complete the square

24 x + 6 x^{2} : 6 (x + 2)^{2} - 24

= \int \frac{1}{6 ( x + 2 ) ^{2} - 24} d x

Apply u-substitution

: \int \frac{1}{6 u ^{2} - 24} d u

= \int \frac{1}{6 u ^{2} - 24} d u

Apply Trigonometric Substitution

: \int \frac{sec ( v )}{6} d v

= \int \frac{sec ( v )}{6} d v

Take the constant out:

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

= \frac{1}{6} \cdot \int sec (v) d v

Use the common integral:

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

= \frac{1}{6} ln ∣ tan (v) + sec (v) ∣

Substitute back

= \frac{1}{6} ln tan (arcsec (\frac{6}{2 6} (x + 2))) + sec (arcsec (\frac{6}{2 6} (x + 2)))

Simplify

\frac{1}{6} ln tan (arcsec (\frac{6}{2 6} (x + 2))) + sec (arcsec (\frac{6}{2 6} (x + 2))) : \frac{1}{6} ln (\frac{x ^{2} + 4 x + x + 2}{2})

= \frac{1}{6} ln (\frac{x ^{2} + 4 x + x + 2}{2})

Add a constant to the solution

= \frac{1}{6} ln (\frac{x ^{2} + 4 x + x + 2}{2}) + C

\int \frac{7}{7 x ^{2} + 14 x + 5} d x

x \to 0 lim (\frac{1}{x})

f (x) = \frac{tan ( x ) - 1}{sec ( x )}

\frac{\partial}{\partial y} (e^{y} \cdot cos (x))

\frac{( s + 3 )}{s ^{2} + 2 s + 1}

What is the integral of 1/(sqrt(24x+6x^2)) ?
The integral of 1/(sqrt(24x+6x^2)) is 1/(sqrt(6))ln((|sqrt(x^2+4x)+x+2|)/2)+C