What is the integral of 1/(sqrt(48x+6x^2)) ?

The integral of 1/(sqrt(48x+6x^2)) is 1/(sqrt(6))ln((|sqrt(x^2+8x)+x+4|)/4)+C

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integral of 1/(sqrt(48x+6x^2))

\int \frac{1}{48 x + 6 x ^{2}} d x

\frac{1}{6} ln (\frac{x ^{2} + 8 x + x + 4}{4}) + C

Solution steps

\int \frac{1}{48 x + 6 x ^{2}} d x

Complete the square

48 x + 6 x^{2} : 6 (x + 4)^{2} - 96

= \int \frac{1}{6 ( x + 4 ) ^{2} - 96} d x

Apply u-substitution

: \int \frac{1}{6 u ^{2} - 96} d u

= \int \frac{1}{6 u ^{2} - 96} 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}{4 6} (x + 4))) + sec (arcsec (\frac{6}{4 6} (x + 4)))

Simplify

\frac{1}{6} ln tan (arcsec (\frac{6}{4 6} (x + 4))) + sec (arcsec (\frac{6}{4 6} (x + 4))) : \frac{1}{6} ln (\frac{x ^{2} + 8 x + x + 4}{4})

= \frac{1}{6} ln (\frac{x ^{2} + 8 x + x + 4}{4})

Add a constant to the solution

= \frac{1}{6} ln (\frac{x ^{2} + 8 x + x + 4}{4}) + C

x \to 6 lim (8 x^{2} - 7 x + 1)

\frac{d}{d x} (3 cos (- x^{3} + 3 x + 1))

\int x - \frac{5 - 25 - 4 x}{2} d x

\int \frac{1}{n + 3}

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

What is the integral of 1/(sqrt(48x+6x^2)) ?
The integral of 1/(sqrt(48x+6x^2)) is 1/(sqrt(6))ln((|sqrt(x^2+8x)+x+4|)/4)+C