What is the integral of x^2(x-4)^{1/2} ?

The integral of x^2(x-4)^{1/2} is 2(1/7 (x-4)^{7/2}+8/5 (x-4)^{5/2}+16/3 (x-4)^{3/2})+C

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integral of x^2(x-4)^{1/2}

integral

x^{2} (x - 4)^{\frac{1}{2}}

2 (\frac{1}{7} (x - 4)^{\frac{7}{2}} + \frac{8}{5} (x - 4)^{\frac{5}{2}} + \frac{16}{3} (x - 4)^{\frac{3}{2}}) + C

Solution steps

\int x^{2} (x - 4)^{\frac{1}{2}} d x

Apply u-substitution

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

Expand

(u^{2} + 4)^{2} u^{2} : u^{6} + 8 u^{4} + 16 u^{2}

= 2 \cdot \int u^{6} + 8 u^{4} + 16 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 u^{6} d u + \int 8 u^{4} d u + \int 16 u^{2} d u)

\int u^{6} d u = \frac{u ^{7}}{7}

\int 8 u^{4} d u = \frac{8 u ^{5}}{5}

\int 16 u^{2} d u = \frac{16 u ^{3}}{3}

= 2 (\frac{u ^{7}}{7} + \frac{8 u ^{5}}{5} + \frac{16 u ^{3}}{3})

Substitute back

u = (x - 4)^{\frac{1}{2}}

= 2 \frac{( ( x - 4 ) ^{\frac{1}{2}} ) ^{7}}{7} + \frac{8 ( ( x - 4 ) ^{\frac{1}{2}} ) ^{5}}{5} + \frac{16 ( ( x - 4 ) ^{\frac{1}{2}} ) ^{3}}{3}

Simplify

2 \frac{( ( x - 4 ) ^{\frac{1}{2}} ) ^{7}}{7} + \frac{8 ( ( x - 4 ) ^{\frac{1}{2}} ) ^{5}}{5} + \frac{16 ( ( x - 4 ) ^{\frac{1}{2}} ) ^{3}}{3} : 2 (\frac{1}{7} (x - 4)^{\frac{7}{2}} + \frac{8}{5} (x - 4)^{\frac{5}{2}} + \frac{16}{3} (x - 4)^{\frac{3}{2}})

= 2 (\frac{1}{7} (x - 4)^{\frac{7}{2}} + \frac{8}{5} (x - 4)^{\frac{5}{2}} + \frac{16}{3} (x - 4)^{\frac{3}{2}})

Add a constant to the solution

= 2 (\frac{1}{7} (x - 4)^{\frac{7}{2}} + \frac{8}{5} (x - 4)^{\frac{5}{2}} + \frac{16}{3} (x - 4)^{\frac{3}{2}}) + C

in t e g r a l x^{2} \cdot (e^{x})^{2}

in t e g r a l e^{x} - 2^{x}

in t e g r a l \frac{x}{( x ^{2} - 1 )}

in t e g r a l \frac{1}{( c - x )}

in t e g r a l \frac{1}{1 + sin ^{2} ( x )}

What is the integral of x^2(x-4)^{1/2} ?
The integral of x^2(x-4)^{1/2} is 2(1/7 (x-4)^{7/2}+8/5 (x-4)^{5/2}+16/3 (x-4)^{3/2})+C