What is the integral of (2x+1)sqrt(x-5) ?

The integral of (2x+1)sqrt(x-5) is 2(2/5 (x-5)^{5/2}+11/3 (x-5)^{3/2})+C

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integral of (2x+1)sqrt(x-5)

\int (2 x + 1) x - 5 d x

2 (\frac{2}{5} (x - 5)^{\frac{5}{2}} + \frac{11}{3} (x - 5)^{\frac{3}{2}}) + C

Solution steps

\int (2 x + 1) x - 5 d x

Apply u-substitution

: \int 2 u^{2} (2 u^{2} + 11) d u

= \int 2 u^{2} (2 u^{2} + 11) d u

Take the constant out:

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

= 2 \cdot \int u^{2} (2 u^{2} + 11) d u

Expand

u^{2} (2 u^{2} + 11) : 2 u^{4} + 11 u^{2}

= 2 \cdot \int 2 u^{4} + 11 u^{2} 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 (\int 2 u^{4} d u + \int 11 u^{2} d u)

\int 2 u^{4} d u = \frac{2 u ^{5}}{5}

\int 11 u^{2} d u = \frac{11 u ^{3}}{3}

= 2 (\frac{2 u ^{5}}{5} + \frac{11 u ^{3}}{3})

Substitute back

u = x - 5

= 2 (\frac{2 ( x - 5 ) ^{5}}{5} + \frac{11 ( x - 5 ) ^{3}}{3})

Simplify

2 (\frac{2 ( x - 5 ) ^{5}}{5} + \frac{11 ( x - 5 ) ^{3}}{3}) : 2 (\frac{2}{5} (x - 5)^{\frac{5}{2}} + \frac{11}{3} (x - 5)^{\frac{3}{2}})

= 2 (\frac{2}{5} (x - 5)^{\frac{5}{2}} + \frac{11}{3} (x - 5)^{\frac{3}{2}})

Add a constant to the solution

= 2 (\frac{2}{5} (x - 5)^{\frac{5}{2}} + \frac{11}{3} (x - 5)^{\frac{3}{2}}) + C

∣ 4 x - 1 ∣, x^{2} - 5

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

x \to 4 lim (7)

x \to 0 lim (\frac{1}{8} + x - (\frac{1}{8}))

\int (5 x^{2} + 3 x - 10) d x

What is the integral of (2x+1)sqrt(x-5) ?
The integral of (2x+1)sqrt(x-5) is 2(2/5 (x-5)^{5/2}+11/3 (x-5)^{3/2})+C