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

The integral of 1/((x+1)sqrt(x^2+2x-3)) is 1/2 arcsec(1/2 (x+1))+C

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

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

\frac{1}{2} arcsec (\frac{1}{2} (x + 1)) + C

Solution steps

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

Apply u-substitution

: \int \frac{1}{u u ^{2} - 4} d u

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

Apply Trigonometric Substitution

: \int \frac{1}{2} d v

= \int \frac{1}{2} d v

Take the constant out:

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

= \frac{1}{2} \cdot \int 1 d v

Integral of a constant:

\int a d x = a x

= \frac{1}{2} \cdot 1 \cdot v

Substitute back

= \frac{1}{2} \cdot 1 \cdot arcsec (\frac{1}{2} (x + 1))

Simplify

= \frac{1}{2} arcsec (\frac{1}{2} (x + 1))

Add a constant to the solution

= \frac{1}{2} arcsec (\frac{1}{2} (x + 1)) + C

\int (\frac{1 + cos ( 2 x )}{2})^{2} d x

cos (ln (t))

x \frac{d y}{d x} + 4 y = x^{- 3} e^{x}

\int \frac{x}{2} d x

\frac{\partial}{\partial x} (2 x^{3} y + 3 x y^{2})

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