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Given the integral $\int_0^4 x^3(x^2 + 1)^{-\frac{1}{2}}dx$, I have tried to choose $u = x^2 + 1, du = 2x\space dx$, thus being left with the integral $\int_1^{17} \frac{1}{2}(du)^ 3\space u^{-\frac{1}{2}}$

Is there a simple way to solve this, or do I need to find a more suitable method?

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  • $\begingroup$ If $u = x^2+1$ then $x^2 = u-1$ and $du = 2xdx$. So, you can write $x^3(x^2+1)^{-1/2}$dx as $x^2 (x^2+1)^{-1/2} x dx = \frac{1}{2}(u-1)\sqrt{u}du$. $\endgroup$
    – Tom
    Oct 31 '13 at 12:05
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    $\begingroup$ @Tom: Make it an answer? $\endgroup$ Oct 31 '13 at 12:06
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Let $u = x^2 +1$. Then we have both $x^2 = u-1$ and $\frac{1}{2} du = xdx$. Rewriting the integrand as $x^3(x^2+1)^{-1/2} dx = x^2 (x^2+1)^{-1/2} \,x\,dx = \frac{1}{2}(u-1)(u)^{-1/2}\,du.$ So, your $u$-sub was right on, just needed to take care of that other term in a reasonable way! Your integral should reduce to $$ \int_0^4 x^3(x^2+1)^{-1/2} \,dx = \frac{1}{2}\int_1^{17}\frac{u-1}{\sqrt{u}} \,du $$

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another way to answer this integral attention when $u=x^2+1$, $u\ge 0$ note that lower limit is $0$ that's not possible because $0=x^2+1 \rightarrow x=?$

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