What is the integral of 2tsqrt(3-4t) ?

The integral of 2tsqrt(3-4t) is -1/10 (2t+1)(-4t+3)^{3/2}+C

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integral of 2tsqrt(3-4t)

\int 2 t 3 - 4 t d t

- \frac{1}{10} (2 t + 1) (- 4 t + 3)^{\frac{3}{2}} + C

Solution steps

\int 2 t 3 - 4 t d t

Take the constant out:

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

= 2 \cdot \int t 3 - 4 t d t

Apply u-substitution

: \int \frac{u ( u - 3 )}{16} d u

= 2 \cdot \int \frac{u ( u - 3 )}{16} d u

Take the constant out:

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

= 2 \cdot \frac{1}{16} \cdot \int u (u - 3) d u

Expand

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

= 2 \cdot \frac{1}{16} \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}{16} (\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}{16} (\frac{2}{5} u^{\frac{5}{2}} - 2 u^{\frac{3}{2}})

Substitute back

u = 3 - 4 t

= 2 \cdot \frac{1}{16} (\frac{2}{5} (3 - 4 t)^{\frac{5}{2}} - 2 (3 - 4 t)^{\frac{3}{2}})

Simplify

2 \cdot \frac{1}{16} (\frac{2}{5} (3 - 4 t)^{\frac{5}{2}} - 2 (3 - 4 t)^{\frac{3}{2}}) : - \frac{1}{10} (2 t + 1) (- 4 t + 3)^{\frac{3}{2}}

= - \frac{1}{10} (2 t + 1) (- 4 t + 3)^{\frac{3}{2}}

Add a constant to the solution

= - \frac{1}{10} (2 t + 1) (- 4 t + 3)^{\frac{3}{2}} + C

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What is the integral of 2tsqrt(3-4t) ?
The integral of 2tsqrt(3-4t) is -1/10 (2t+1)(-4t+3)^{3/2}+C