What is the integral of 1/(sqrt(3x^2+6x-4)) ?

The integral of 1/(sqrt(3x^2+6x-4)) is 1/(sqrt(3))ln|sqrt((3x^2+6x-4)/7)+sqrt(3/7)(x+1)|+C

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integral of 1/(sqrt(3x^2+6x-4))

\int \frac{1}{3 x ^{2} + 6 x - 4} d x

\frac{1}{3} ln \frac{3 x ^{2} + 6 x - 4}{7} + \frac{3}{7} (x + 1) + C

Solution steps

\int \frac{1}{3 x ^{2} + 6 x - 4} d x

Complete the square

3 x^{2} + 6 x - 4 : 3 (x + 1)^{2} - 7

= \int \frac{1}{3 ( x + 1 ) ^{2} - 7} d x

Apply u-substitution

: \int \frac{1}{3 u ^{2} - 7} d u

= \int \frac{1}{3 u ^{2} - 7} d u

Apply Trigonometric Substitution

: \int \frac{sec ( v )}{3} d v

= \int \frac{sec ( v )}{3} d v

Take the constant out:

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

= \frac{1}{3} \cdot \int sec (v) d v

Use the common integral:

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

= \frac{1}{3} ln ∣ tan (v) + sec (v) ∣

Substitute back

= \frac{1}{3} ln tan (arcsec (\frac{3}{7} (x + 1))) + sec (arcsec (\frac{3}{7} (x + 1)))

Simplify

\frac{1}{3} ln tan (arcsec (\frac{3}{7} (x + 1))) + sec (arcsec (\frac{3}{7} (x + 1))) : \frac{1}{3} ln \frac{3 x ^{2} + 6 x - 4}{7} + \frac{3}{7} (x + 1)

= \frac{1}{3} ln \frac{3 x ^{2} + 6 x - 4}{7} + \frac{3}{7} (x + 1)

Add a constant to the solution

= \frac{1}{3} ln \frac{3 x ^{2} + 6 x - 4}{7} + \frac{3}{7} (x + 1) + C

n = 1 \sum \infty \frac{1}{n + 5}

x \to 1 lim (\frac{2 x ^{2}}{x - 1})

\frac{d y}{d x} + 2 y = 3 x

y^{^{''}} - 10 y^{^{'}} + 26 y = 0

(1 + x) y^{^{'}} + y = x

What is the integral of 1/(sqrt(3x^2+6x-4)) ?
The integral of 1/(sqrt(3x^2+6x-4)) is 1/(sqrt(3))ln|sqrt((3x^2+6x-4)/7)+sqrt(3/7)(x+1)|+C