What is the integral of 2xsqrt(y-1) ?

The integral of 2xsqrt(y-1) is x^2sqrt(-1+y)+C

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integral of 2xsqrt(y-1)

\int 2 x y - 1 d x

x^{2} - 1 + y + C

Solution steps

\int 2 x y - 1 d x

Take the constant out:

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

= 2 - 1 + y \cdot \int x d x

Apply the Power Rule

: \frac{x ^{2}}{2}

= 2 - 1 + y \frac{x ^{2}}{2}

Simplify

2 - 1 + y \frac{x ^{2}}{2} : x^{2} - 1 + y

= x^{2} - 1 + y

Add a constant to the solution

= x^{2} - 1 + y + C

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

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\int \frac{x + 3}{x ^{2} + 4 x + 5} d x

\frac{1}{x 1 - x} d x

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