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

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

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

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

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

Solution steps

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

Complete the square

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

= \int \frac{1}{( x + \frac{1}{2} ) ^{2} - \frac{5}{4}} d x

Apply u-substitution

: \int \frac{2}{4 u ^{2} - 5} d u

= \int \frac{2}{4 u ^{2} - 5} d u

Take the constant out:

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

= 2 \cdot \int \frac{1}{4 u ^{2} - 5} d u

Apply Trigonometric Substitution

: \int \frac{sec ( v )}{2} d v

= 2 \cdot \int \frac{sec ( v )}{2} d v

Take the constant out:

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

= 2 \cdot \frac{1}{2} \cdot \int sec (v) d v

Use the common integral:

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

= 2 \cdot \frac{1}{2} ln ∣ tan (v) + sec (v) ∣

Substitute back

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

Simplify

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

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

Add a constant to the solution

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

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