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

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

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 a substitution, but why does my solution not work?

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    $\begingroup$ Don't forget your $+C$! $\endgroup$ Commented May 13, 2019 at 12:53
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    $\begingroup$ 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. $\endgroup$ Commented May 13, 2019 at 12:54
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    $\begingroup$ 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? $\endgroup$ Commented May 13, 2019 at 12:54
  • $\begingroup$ Oops @CameronWilliams was faster than I. $\endgroup$ Commented May 13, 2019 at 12:55
  • $\begingroup$ 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. $\endgroup$ Commented May 13, 2019 at 13:31

3 Answers 3

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

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

or

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

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    $\begingroup$ Thank you Ak19, great solution! $\endgroup$ Commented May 13, 2019 at 13:32
  • $\begingroup$ You're welcome:) $\endgroup$ Commented May 13, 2019 at 13:33
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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 $$\frac{ 2(x-1)^{\frac{3}{2}}}{15}[3x+2] +C $$

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    $\begingroup$ Thank you Vedant Chourney, your solution is the third possible one! $\endgroup$ Commented May 13, 2019 at 13:34
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I will explain the logic behind the substitution $t=\sqrt{x-1}$. The given integral can be recognized as Chebyshev's differential binomial:

$$\int x^m(ax^n+b)^p\mathrm dx$$

Cases when this integral is elementary:

$1.$ If $p\in\mathbb Z^+$, expanding using the binomial theorem.

$2.$ If $p\in\mathbb Z^-$ and $m,n\in\mathbb Q$, substitute $x=t^k$ where $k$ is the LCM of the denominators of $m$ and $n$.

$3.$ If $\frac{m+1}n\in\mathbb Z$ and $p\in\mathbb Q$, substitute $t=(ax^n+b)^\frac1k$ where $k$ is the denominator of $p$.

$4.$ If $\frac{m+1}n+p\in\mathbb Z$ and $p\in\mathbb Q$, substitute $t=(a+bx^{-n})^\frac1k$ where $k$ is the denominator of $p$.

Your integral falls into case $3$ as mentioned above, hence it is feasible to proceed forward with the substitution $t=\sqrt{x-1}$; it converts the integrand into a polynomial in $t$.

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