What is the integral of x^2sqrt(x-3) ?

The integral of x^2sqrt(x-3) is 2/7 (x-3)^{7/2}+12/5 (x-3)^{5/2}+6(x-3)^{3/2}+C

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integral of x^2sqrt(x-3)

\int x^{2} x - 3 d x

\frac{2}{7} (x - 3)^{\frac{7}{2}} + \frac{12}{5} (x - 3)^{\frac{5}{2}} + 6 (x - 3)^{\frac{3}{2}} + C

Solution steps

\int x^{2} x - 3 d x

Apply u-substitution

: \int (u + 3)^{2} u d u

= \int (u + 3)^{2} u d u

Expand

(u + 3)^{2} u : u^{\frac{5}{2}} + 6 u^{\frac{3}{2}} + 9 u

= \int u^{\frac{5}{2}} + 6 u^{\frac{3}{2}} + 9 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

= \int u^{\frac{5}{2}} d u + \int 6 u^{\frac{3}{2}} d u + \int 9 u d u

\int u^{\frac{5}{2}} d u = \frac{2}{7} u^{\frac{7}{2}}

\int 6 u^{\frac{3}{2}} d u = \frac{12}{5} u^{\frac{5}{2}}

\int 9 u d u = 6 u^{\frac{3}{2}}

= \frac{2}{7} u^{\frac{7}{2}} + \frac{12}{5} u^{\frac{5}{2}} + 6 u^{\frac{3}{2}}

Substitute back

u = x - 3

= \frac{2}{7} (x - 3)^{\frac{7}{2}} + \frac{12}{5} (x - 3)^{\frac{5}{2}} + 6 (x - 3)^{\frac{3}{2}}

Add a constant to the solution

= \frac{2}{7} (x - 3)^{\frac{7}{2}} + \frac{12}{5} (x - 3)^{\frac{5}{2}} + 6 (x - 3)^{\frac{3}{2}} + C

\frac{d}{d x} (x^{3} - 2 x^{2} + 4)

x \to 0 lim (\frac{3 sin ( x )}{x})

\frac{1}{2 + x}

\int \frac{( x ^{2} + 5 x + 4 )}{x ^{2} - 2 x + 1} d x

What is the integral of x^2sqrt(x-3) ?
The integral of x^2sqrt(x-3) is 2/7 (x-3)^{7/2}+12/5 (x-3)^{5/2}+6(x-3)^{3/2}+C