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I am trying to evaluate this integral and faced with the following problem: $$\int_{1}^{2}(\sqrt{x^{2}+1}\cdot x) dx$$ what I tried to do is:
set $\sqrt{x^2+1}=t$ tried to make derivative but not success.
any suggestions?
thanks!

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  • $\begingroup$ take x = tan(t) $\endgroup$ – Sumedh Sep 3 '13 at 11:26
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    $\begingroup$ No substitution: $$\int \sqrt {x^2+1}x\,\mathrm dx=\dfrac 12\int 2x(x^2+1)^{1/2}\mathrm dx\overset{f(x)=x^2+1}{=}\dfrac 12 \int f'(x)(f(x))^{1/2}\mathrm dx=\\ =\dfrac {1}{2} \int \left(\dfrac 23(f(x))^{3/2}\right)'\mathrm dx=\dfrac{(f(x))^{3/2}}{3}=\dfrac{(x^2+1)^{3/2}}{3}.$$ Now compute the definite integral. $\endgroup$ – Git Gud Sep 3 '13 at 11:28
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Setting $$ t=x^2+1, $$ we get \begin{eqnarray} \int_1^2 x\sqrt{x^2+1}\,dx&=&\frac12\int_2^5\sqrt{t}\,dt=\frac13\left[t^{3/2}\right]_2^5=\frac{5\sqrt{5}-2\sqrt{2}}{3}. \end{eqnarray}

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Try making the substitution $t = 1+x^2$ which has $dt = 2x~dx$. Often times it's easier to use just the stuff under the radical when doing substitutions so you don't have to deal as much with the fact that the derivative of $\sqrt x$ is $\frac{-1}{\sqrt x}$.

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Hint: The square root is going to give you problems. Try $t = x^2 + 1$.

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