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Question Number 182771 by liuxinnan last updated on 14/Dec/22

$$\int {}_0^r\frac{1}{R^2-r^2}dr=?$$

Commented by Frix last updated on 14/Dec/22

$$\mathrm{Error}:\mathrm{dependent}\mathrm{borders}\mathrm{are}\mathrm{not}\mathrm{allowed}.$$

Commented by mr W last updated on 14/Dec/22

$$whynot?$$$${\int }_0^r\frac{1}{R^2-r^2}dr$$$$={\int }_0^r\frac{1}{R^2-x^2}dx$$$$={\int }_0^r\frac{1}{R^2-y^2}dy$$$$={\int }_0^r\frac{1}{R^2-t^2}dt$$$$={\int }_0^r\frac{1}{R^2-{\lambda }^2}d\lambda$$$$...$$

Commented by Frix last updated on 14/Dec/22

$${\mathrm{I}}^{\prime }\mathrm{m}\mathrm{not}\mathrm{sure},\mathrm{it}\mathrm{might}\mathrm{just}\mathrm{be}\mathrm{a}\mathrm{formal}$$$$\mathrm{problem}...$$

Commented by Frix last updated on 14/Dec/22

$$\mathrm{My}\mathrm{thoughts}:$$$$\int f\left(x\right)dx=F\left(x\right)+C$$$$\mathrm{gives}\mathrm{the}\mathrm{antiderivative}$$$$\underset{a}{\stackrel{b}{\int }}f\left(x\right)dx=F\left(b\right)-F\left(a\right)$$$$\mathrm{gives}\mathrm{the}\mathrm{area}\mathrm{between}f\left(x\right)\mathrm{and}\mathrm{the}x-\mathrm{axis}$$$$\underset{a}{\stackrel{x}{\int }}f\left(x\right)dx=F\left(x\right)-F\left(a\right)$$$$\mathrm{gives}what?\mathrm{I}\mathrm{never}\mathrm{needed}\mathrm{this}...\mathrm{anyway}$$$${\mathrm{it}}^{\prime }\mathrm{s}\mathrm{the}\mathrm{first}\mathrm{case}\mathrm{with}C=-F\left(a\right)$$

Commented by mr W last updated on 14/Dec/22

$$\underset{a}{\stackrel{b}{\int }}f\left(x\right)dx=F\left(b\right)-F\left(a\right)$$$$youcanreplacebwithtorxasyou$$$$likeandget$$$$\underset{a}{\stackrel{t}{\int }}f\left(x\right)dx=F\left(t\right)-F\left(a\right)$$$$\underset{a}{\stackrel{x}{\int }}f\left(x\right)dx=F\left(x\right)-F\left(a\right)$$

Commented by mr W last updated on 14/Dec/22

$$\underset{a}{\stackrel{x}{\int }}f\left(x\right)dx=F\left(x\right)-F\left(a\right)givesthearea$$$$underthecurvey=f\left(x\right)fromx=a$$$$tox=x.$$

Commented by mr W last updated on 14/Dec/22

Answered by mr W last updated on 14/Dec/22

$${\int }_0^r\frac{1}{R^2-r^2}dr$$$$=\frac{1}{2R}{\int }_0^r(\frac{1}{R-r}+\frac{1}{R+r})dr$$$$=\frac{1}{2R}{\left[\ln \frac{R+r}{R-r}\right]}_0^r$$$$=\frac{1}{2R}\ln \frac{R+r}{R-r}$$

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