Question

\[\int\limits_{1}^{6} x^2/\sqrt{x+3}\]

Answer 1

how do i evaluate that integral

Answer 2

int by subs, let u=x+3

Answer 3

\[\int_1^6\frac{x^2}{\sqrt{x+3}}\text dx\]

\(u=x+3\)

\(\frac{\text du}{\text dx}=1\)

\(\text du=\text dx\)

\[=\int_1^6\frac{(u-3)^2}{\sqrt{u}}\text du\]

Answer 4

\[=\int_1^6\frac{u^2-6u+9}{\sqrt{u}}\text du\]
\[=\int_1^6\frac{u^2}{\sqrt{u}}\text du-6\int_1^6\frac{u}{\sqrt{u}}\text du+9\int_1^6\frac{1}{\sqrt{u}}\text du\]

Answer 5

^?

Answer 6

no?