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

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

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

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

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

Solution steps

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

Complete the square

x^{2} + 3 x - 4 : (x + \frac{3}{2})^{2} - \frac{25}{4}

= \int \frac{1}{( x + \frac{3}{2} ) ^{2} - \frac{25}{4}} d x

Apply u-substitution

: \int \frac{2}{4 u ^{2} - 25} d u

= \int \frac{2}{4 u ^{2} - 25} 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} - 25} 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{3}{2}))) + sec (arcsec (\frac{2}{5} (x + \frac{3}{2})))

Simplify

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

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

Add a constant to the solution

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

y^{^{''}} + 6 y^{^{'}} + 9 y = 0, y (0) = 2, y^{^{'}} (0) = - 7

\int sin (x) + sec^{2} (x) + sec (x) tan (x) d x

\int \frac{3}{x} + x^{3} d x

\frac{\partial}{\partial x} (4 y cos (x))

\frac{d x}{d t} = sin (3 t) cos (3 t)

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