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How can I compute

$$\int_0^1 \frac{x^3t}{(x^2+t^2)^2} \, \mathrm{dt}$$

by hand?

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  • $\begingroup$ Hint: $u=x^2+t^2$. $\endgroup$ – Ghartal Jun 25 '16 at 13:02
  • $\begingroup$ Pretty close to elementary, isn't it? Seeing the "$x^2+ t^2$" in the denominator and "$tdt$" in the numerator, my first thought would be "try $u= x^2+ t^2$. You realize, do you not, that "x" is a constant here? What is "du"? $\endgroup$ – user247327 Jun 25 '16 at 13:05
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Hint: $$\int \frac{x^3t}{(x^2+t^2)^2}dt$$ $$=x^3\int \frac{t}{(x^2+t^2)^2}dt$$ $$u=(x^2+t^2)\Rightarrow du=2tdt$$ $$=x^3\frac{1}{2}\int \frac{1}{u^2}du$$ $$=x^3\frac{1}{2}\int u^{-2}du$$ It's pretty simple from here, you can simply use the power rule, substitute $u$ back in and simplify to have the indefinite integral. From there just compute the bounds.

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  • $\begingroup$ Thanks, now that I look at it it was actually pretty easy. The x² confused me. $\endgroup$ – Staki42 Jun 25 '16 at 13:41

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