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

The integral of 1/(sqrt(x+6x^2)) is 1/(sqrt(6))ln|sqrt(144x^2+24x)+12x+1|+C

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

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

\frac{1}{6} ln 144 x^{2} + 24 x + 12 x + 1 + C

Solution steps

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

Complete the square

x + 6 x^{2} : 6 (x + \frac{1}{12})^{2} - \frac{1}{24}

= \int \frac{1}{6 ( x + \frac{1}{12} ) ^{2} - \frac{1}{24}} d x

Apply u-substitution

: \int \frac{2 6}{144 u ^{2} - 1} d u

= \int \frac{2 6}{144 u ^{2} - 1} d u

Take the constant out:

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

= 26 \cdot \int \frac{1}{144 u ^{2} - 1} d u

Apply Trigonometric Substitution

: \int \frac{sec ( v )}{12} d v

= 26 \cdot \int \frac{sec ( v )}{12} d v

Take the constant out:

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

= 26 \frac{1}{12} \cdot \int sec (v) d v

Use the common integral:

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

= 26 \frac{1}{12} ln ∣ tan (v) + sec (v) ∣

Substitute back

= 26 \frac{1}{12} ln tan (arcsec (12 (x + \frac{1}{12}))) + sec (arcsec (12 (x + \frac{1}{12})))

Simplify

26 \frac{1}{12} ln tan (arcsec (12 (x + \frac{1}{12}))) + sec (arcsec (12 (x + \frac{1}{12}))) : \frac{1}{6} ln 144 x^{2} + 24 x + 12 x + 1

= \frac{1}{6} ln 144 x^{2} + 24 x + 12 x + 1

Add a constant to the solution

= \frac{1}{6} ln 144 x^{2} + 24 x + 12 x + 1 + C

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What is the integral of 1/(sqrt(x+6x^2)) ?
The integral of 1/(sqrt(x+6x^2)) is 1/(sqrt(6))ln|sqrt(144x^2+24x)+12x+1|+C