What is the integral of x^2sqrt(7+x) ?

The integral of x^2sqrt(7+x) is 2/7 (7+x)^{7/2}-28/5 (7+x)^{5/2}+98/3 (7+x)^{3/2}+C

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integral of x^2sqrt(7+x)

\int x^{2} 7 + x d x

\frac{2}{7} (7 + x)^{\frac{7}{2}} - \frac{28}{5} (7 + x)^{\frac{5}{2}} + \frac{98}{3} (7 + x)^{\frac{3}{2}} + C

Solution steps

\int x^{2} 7 + x d x

Apply u-substitution

: \int (u - 7)^{2} u d u

= \int (u - 7)^{2} u d u

Expand

(u - 7)^{2} u : u^{\frac{5}{2}} - 14 u^{\frac{3}{2}} + 49 u

= \int u^{\frac{5}{2}} - 14 u^{\frac{3}{2}} + 49 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 14 u^{\frac{3}{2}} d u + \int 49 u d u

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

\int 14 u^{\frac{3}{2}} d u = \frac{28}{5} u^{\frac{5}{2}}

\int 49 u d u = \frac{98}{3} u^{\frac{3}{2}}

= \frac{2}{7} u^{\frac{7}{2}} - \frac{28}{5} u^{\frac{5}{2}} + \frac{98}{3} u^{\frac{3}{2}}

Substitute back

u = 7 + x

= \frac{2}{7} (7 + x)^{\frac{7}{2}} - \frac{28}{5} (7 + x)^{\frac{5}{2}} + \frac{98}{3} (7 + x)^{\frac{3}{2}}

Add a constant to the solution

= \frac{2}{7} (7 + x)^{\frac{7}{2}} - \frac{28}{5} (7 + x)^{\frac{5}{2}} + \frac{98}{3} (7 + x)^{\frac{3}{2}} + C

\int t^{2} 7 t - 5 d t

x \to 0 + lim (ln ∣ x ∣)

4 sec^{2} (x)

\frac{x ^{2}}{7 + 4 x}

x \to \infty lim (\frac{1}{2})

What is the integral of x^2sqrt(7+x) ?
The integral of x^2sqrt(7+x) is 2/7 (7+x)^{7/2}-28/5 (7+x)^{5/2}+98/3 (7+x)^{3/2}+C