How to calculate $$\int_0^1 \frac{t \, \mathrm{d}t}{\sqrt{(t-x)^2+y^2+z^2}}$$ Here, $x,y,z$ are fixed.
How to calculate $\int_0^1 \frac{t dt}{\sqrt{(t-x)^2+y^2+z^2}}$?
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calculus
integration
1 Answers
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Let $u=t-x$, we get $$I=\int \frac{t }{\sqrt{(t-x)^2+y^2+z^2}}\, \mathrm{d}t=\int \frac{u+x}{\sqrt{u^{2}+y^{2}+z^{2}}}\, \mathrm{d}u$$ then let $u=\sqrt{y^{2}+z^{2}}\tan m$ \begin{align*} I&=\int \sec m\left ( \sqrt{y^{2}+z^{2}}\tan m+x \right )\, \mathrm{d}m\\ &=\sqrt{y^{2}+z^{2}}\int \tan m\sec m\, \mathrm{d}m+x\int \sec m\, \mathrm{d}m\\ &=\sqrt{y^{2}+z^{2}}\sec m+x\ln\left | \tan m+\sec m \right |+C \end{align*} Hope you can take it from here.