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

The integral of (x+1)sqrt(2x-1) is 1/3 (x+1)(2x-1)^{3/2}-1/15 (2x-1)^{5/2}+C

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integral of (x+1)sqrt(2x-1)

\int (x + 1) 2 x - 1 d x

\frac{1}{3} (x + 1) (2 x - 1)^{\frac{3}{2}} - \frac{1}{15} (2 x - 1)^{\frac{5}{2}} + C

Solution steps

\int (x + 1) 2 x - 1 d x

Apply Integration By Parts

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

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

\int \frac{1}{3} (2 x - 1)^{\frac{3}{2}} d x = \frac{1}{15} (2 x - 1)^{\frac{5}{2}}

= \frac{1}{3} (x + 1) (2 x - 1)^{\frac{3}{2}} - \frac{1}{15} (2 x - 1)^{\frac{5}{2}}

Add a constant to the solution

= \frac{1}{3} (x + 1) (2 x - 1)^{\frac{3}{2}} - \frac{1}{15} (2 x - 1)^{\frac{5}{2}} + C

\frac{d x}{d t} = x^{2}, x (0) = 1

\int - 6 x^{- 7} d x

\frac{d y}{d x} + (\frac{1 - y ^{2}}{1 - x ^{2}})^{\frac{1}{2}} = 0

\frac{d y}{d x} = 9 y

G (t) = \frac{1 - 3 t}{4 + t}

What is the integral of (x+1)sqrt(2x-1) ?
The integral of (x+1)sqrt(2x-1) is 1/3 (x+1)(2x-1)^{3/2}-1/15 (2x-1)^{5/2}+C