What is the integral of 2tsqrt(2t-3) ?

The integral of 2tsqrt(2t-3) is 2/5 (t+1)(2t-3)^{3/2}+C

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integral of 2tsqrt(2t-3)

\int 2 t 2 t - 3 d t

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

Solution steps

\int 2 t 2 t - 3 d t

Take the constant out:

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

= 2 \cdot \int t 2 t - 3 d t

Apply u-substitution

: \int \frac{u ( u + 3 )}{4} d u

= 2 \cdot \int \frac{u ( u + 3 )}{4} d u

Take the constant out:

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

= 2 \cdot \frac{1}{4} \cdot \int u (u + 3) d u

Expand

u (u + 3) : u^{\frac{3}{2}} + 3 u

= 2 \cdot \frac{1}{4} \cdot \int u^{\frac{3}{2}} + 3 u d u

Apply the Sum Rule:

\int f (x) \pm g (x) d x = \int f (x) d x \pm \int g (x) d x

= 2 \cdot \frac{1}{4} (\int u^{\frac{3}{2}} d u + \int 3 u d u)

\int u^{\frac{3}{2}} d u = \frac{2}{5} u^{\frac{5}{2}}

\int 3 u d u = 2 u^{\frac{3}{2}}

= 2 \cdot \frac{1}{4} (\frac{2}{5} u^{\frac{5}{2}} + 2 u^{\frac{3}{2}})

Substitute back

u = 2 t - 3

= 2 \cdot \frac{1}{4} (\frac{2}{5} (2 t - 3)^{\frac{5}{2}} + 2 (2 t - 3)^{\frac{3}{2}})

Simplify

2 \cdot \frac{1}{4} (\frac{2}{5} (2 t - 3)^{\frac{5}{2}} + 2 (2 t - 3)^{\frac{3}{2}}) : \frac{2}{5} (t + 1) (2 t - 3)^{\frac{3}{2}}

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

Add a constant to the solution

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

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