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

The integral of 3/(sqrt(x^2+4x-1)) is 3ln|sqrt(1/5 (x^2+4x-1))+1/(sqrt(5))(x+2)|+C

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

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

3 ln \frac{1}{5} (x^{2} + 4 x - 1) + \frac{1}{5} (x + 2) + C

Solution steps

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

Take the constant out:

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

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

Complete the square

x^{2} + 4 x - 1 : (x + 2)^{2} - 5

= 3 \cdot \int \frac{1}{( x + 2 ) ^{2} - 5} d x

Apply u-substitution

: \int \frac{1}{u ^{2} - 5} d u

= 3 \cdot \int \frac{1}{u ^{2} - 5} d u

Apply Trigonometric Substitution

: \int sec (v) d v

= 3 \cdot \int sec (v) d v

Use the common integral:

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

= 3 ln ∣ tan (v) + sec (v) ∣

Substitute back

= 3 ln tan (arcsec (\frac{1}{5} (x + 2))) + sec (arcsec (\frac{1}{5} (x + 2)))

Simplify

3 ln tan (arcsec (\frac{1}{5} (x + 2))) + sec (arcsec (\frac{1}{5} (x + 2))) : 3 ln \frac{1}{5} (x^{2} + 4 x - 1) + \frac{1}{5} (x + 2)

= 3 ln \frac{1}{5} (x^{2} + 4 x - 1) + \frac{1}{5} (x + 2)

Add a constant to the solution

= 3 ln \frac{1}{5} (x^{2} + 4 x - 1) + \frac{1}{5} (x + 2) + C

\int \frac{10}{2 x + 5} d x

\int 5 e^{x} + 5 d x

\int e^{(- 4 x)} d x

\int 5 e^{x} + 3 d x

What is the integral of 3/(sqrt(x^2+4x-1)) ?
The integral of 3/(sqrt(x^2+4x-1)) is 3ln|sqrt(1/5 (x^2+4x-1))+1/(sqrt(5))(x+2)|+C