Can anyone solve this integral?

Question

What is the integral of

(x^2)((x-7)^1/2)

Answer 1

put x-7 = u dx = du
x =u+7
so integrand is (u+7)^2*u^(1/2)= (u^2 + 14U + 49)* U^(0.5)
= U^(2.5) +14 U ^(1.5) + 49 U ^(0.5) -------(1)


take integrand dx = integrand du

U^n du = U^(n+1)/(n+1)

apply this formula to each of the three terms in eqn (1)

answer is = U^(3.5)/(3.5) + 14 U^(2.5)/(2.5) + 49 U ^(1.5)/1.5

=2U^(7/2)/(7) + 28U^(5/2)/5 + 98 U ^(3/2)/3


now substitute u = x-7

--- thanks!

Answer 2

Im gonna put the word integral inplace of the little squiggly line.

Substitute u=x-7 then set it equal to x, so x=u+7.

Now substitute everything in the inegral for x with u+7

Integral: (u+7)^2 (u+7-7)^1/2

Now simplify..

Integral: (u+7)^2 (u)^1/2

(u+7)(u+7) (u)^1/2 --->(2u^2+14u+49) (u)^1/2

Now multiply u^1/2 into the equation to make it one big equation.

Integral: (u^5/2+14u^3/2+49u^1/2)

Now find the anti-derivative

2/7u^7/2+28/5u^5/2+93/3u^3/2

Answer 3

∫(x^2)((x-7)^1/2) dx: let u = x-7, du = dx
= ∫(u+7)^2.u^(1/2) du
= ∫(u^2 + 14u + 49) u^(1/2) du
= ∫(u^(5/2) + 14u^(3/2) + 49u^(1/2)) du
= u^(7/2) / (7/2) + 14 u^(5/2) / (5/2) + 49 u^(3/2) / (3/2) + c
= (2/7)(x-7)^(7/2) + (28/5)(x-7)^(5/2) + (98/3)(x-7)^(3/2) + c
If you like you can take out a common factor of √(x-7) and expand and simplify the rest; you'd get √(x-7) multiplied by a cubic in x, plus an arbitrary constant. But I don't think this is worth the effort.