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

The integral of (y+1)/(sqrt(y^2+2y-4)) is sqrt(y^2+2y-4)+C

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

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

y^{2} + 2 y - 4 + C

Solution steps

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

Apply u-substitution

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

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

Take the constant out:

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

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

Apply the Power Rule

: 2 u^{\frac{1}{2}}

= \frac{1}{2} \cdot 2 u^{\frac{1}{2}}

Substitute back

u = y^{2} + 2 y - 4

= \frac{1}{2} \cdot 2 (y^{2} + 2 y - 4)^{\frac{1}{2}}

Simplify

\frac{1}{2} \cdot 2 (y^{2} + 2 y - 4)^{\frac{1}{2}} : y^{2} + 2 y - 4

= y^{2} + 2 y - 4

Add a constant to the solution

= y^{2} + 2 y - 4 + C

\int 9 x^{4} d x

x \to 3 - lim (c x^{2} + 2 x)

x^{^{'}} = \frac{1}{x ^{2}}

\frac{d}{d x} (x + \frac{x ^{2}}{2})

\frac{\partial}{\partial x} (5 x + 3 y)

What is the integral of (y+1)/(sqrt(y^2+2y-4)) ?
The integral of (y+1)/(sqrt(y^2+2y-4)) is sqrt(y^2+2y-4)+C