# Indefinite integral of $\int x\sqrt{x-1} \, \mathrm dx$

How can I evaluate the indefinite integral $$\int x\sqrt{x-1} \, \mathrm dx?$$

I tried to calculate it using integration by parts, I get $$\int x\sqrt{x-1} \, \mathrm dx = \frac{2}{3}x(x-1)^{3/2} - \frac{2}{3}\cdot\frac{2}{5}\cdot(x-1)^{5/2}$$

But this is not the correct solution, and I don't understand why.

In the integration by parts formula I used $$f(x)=x$$ and $$g'(x)=(x-1)^{1/2}$$ so $$f'(x)=1$$ and $$g(x)=(x-1)^{3/2}\cdot\frac{2}{3}$$.

What did I do wrong? I know I should use substitutional integral, but why does my solution not work? Thank you

• Don't forget your $+C$! – Cameron Williams May 13 '19 at 12:53
• Your answer is correct though. Take a derivative and you'll see it. You just have it written in a different form that one might expect. That does not make it any less valid. – Cameron Williams May 13 '19 at 12:54
• Except for the Thought-Police-required "+C" - which really is required - your answer is OK. Why do you think your answer is wrong? To check, differentiate your putative anti-derivative. You do get the integrand back, so you're fine....The "+C" can make answers that look different be the same. What is the answer that you expected? – peter a g May 13 '19 at 12:54
• Oops @CameronWilliams was faster than I. – peter a g May 13 '19 at 12:55
• Thank you Cameron Williams :), after some algebraic manipulation I got the same result as I gained by checking in the internet, which was (2/3)*(x-1)^1.5+(6/15)*(x-1)^2.5 And of course, I forgot the C. – TTomi May 13 '19 at 13:31

$$I = \int x\sqrt{x-1}dx = \int (x-1+1)\sqrt{x-1}dx = \int[(x-1)^{3/2} + (x-1)^{1/2}]dx$$

$$I =\frac{2}{5}(x-1)^{5/2} + \frac{2}{3}(x-1)^{3/2} +c$$

or

$$I = \frac{2}{15}\cdot(x-1)^{3/2}(3x+2)+c$$

• Thank you Ak19, great solution! – TTomi May 13 '19 at 13:32
• You're welcome:) – Ak. May 13 '19 at 13:33

Another method can be done by substitution as: Substitite $$(x-1) = u^2$$ That gives $$x = u^2+1$$ And $$dx = 2udu$$ The integral becomes $$\int {2 u^2(u^2+1)} du$$ And its integration is $$\frac {2u^3}{15} [3u^2+5] +C$$ After substituting the value of $$u^2=x-1$$ you get $$\frac{ 2(x-1)^{\frac{3}{2}}}{15}[3x+2] +C$$

• Thank you Vedant Chourney, your solution is the third possible one! – TTomi May 13 '19 at 13:34
• Your welcome :) – Vedant Chourey May 13 '19 at 13:39