What is the integral of (2x+3)/(sqrt(x^2+3x-5)) ?

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

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integral of (2x+3)/(sqrt(x^2+3x-5))

\int \frac{2 x + 3}{x ^{2} + 3 x - 5} d x

2 x^{2} + 3 x - 5 + C

Solution steps

\int \frac{2 x + 3}{x ^{2} + 3 x - 5} d x

Apply u-substitution

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

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

Apply the Power Rule

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

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

Substitute back

u = x^{2} + 3 x - 5

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

Apply radical rule:

= 2 x^{2} + 3 x - 5

Add a constant to the solution

= 2 x^{2} + 3 x - 5 + C

\int ln (\frac{1}{x}) d x

\frac{d y}{d t} = (a \cdot 2) - \frac{y}{20} 2, y (0) = 0

\frac{d y}{d x} = x (4 x + 5)^{\frac{1}{3}}

f (x) = 2 x - x^{2}

\frac{d}{d x} (ln (x^{\frac{1}{x}}))

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