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Evaluate $$\int\sqrt{x^3+x^2}\;dx$$


What I have tried

Using substitution (which I believe was applied incorrectly) I get: $$\frac{(x+2)\sqrt{4x+4}}{4x+4}$$

How can this integral be evaluated?

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  • $\begingroup$ Write the integrand as $x\sqrt{x+1}$ and substitute $u = \sqrt{x+1}$ $\endgroup$ – Prahlad Vaidyanathan Oct 4 '13 at 17:47
  • $\begingroup$ The Maple command $$Student[Calculus1]:-IntTutor(sqrt(x^3+x^2), x) $$ produces the answer $$2/5\, \left( x+1 \right) ^{5/2}-2/3\, \left( x+1 \right) ^{3/2} $$ step by step with explanation. See that link for info. $\endgroup$ – user64494 Oct 4 '13 at 18:00
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For the first integral, suppose that $x\gt 0$. Let $u^2=1+x$. Then $$\sqrt{x^3+x^2}=x\sqrt{1+x}=(u^2-1)(u).$$ Since $2u\,du=dx$, we end up with the integral $$\int (2u)(u^2-1)(u) \,du,$$ which is easy.

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  • 2
    $\begingroup$ Andre Nicolas: WOW you are almost 200K!!!!! $\endgroup$ – user95440 Oct 4 '13 at 18:12
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$$ \int\sqrt{\vphantom{\Large A}x^{3} + x^{2}\,} = \int\left[% \left(x + 1\right)^{3/2} - \left(x + 1\right)^{1/2} \right] = {2 \over 5}\,\left(x + 1\right)^{5/2} - {2 \over 3}\,\left(x + 1\right)^{3/2} + \mbox{constant} $$

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