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Chapter 11: Problem 956

If \(\int\left[\left\\{x^{2012} \tan ^{-1}\left(x^{2012}\right)\right\\} /\left(1+x^{4024}\right)\right] d x=k \tan ^{-1}\left(x^{2012}\right)+c\) (a) \([1 /(2012)]\) (a) \(-[1 /(2012)]\) (c) \([1 /(4024)]\) (d) \(-[1 /(4024)]\)

Short Answer

Expert verified
(a) \([1/(2012)]\)

Step by step solution

01

Recognize the functions for integration by parts

Integration by parts formula is: \[\int u \, dv = uv - \int v \, du\] Here, we can choose the two functions to apply the integration by parts formula: Let u = \(\tan^{-1}(x^{2012})\) Let dv = \(\frac{x^{2012}}{1+x^{4024}}dx\)
02

Calculate du and v

We need to find du and v for using integration by parts formula. Derivative of u = du: \[du = \frac{d}{dx}(\tan^{-1}(x^{2012}))\cdot dx = \frac{2012x^{2011}}{1+x^{4024}}dx\] Now for v, we have: \(v = \int \frac{x^{2012}}{1+x^{4024}}dx\) Notice that the denominator is a perfect square: \((1+x^{2012})^2 = 1+x^{4024}\). Let's substitute: \(x^{2012} = y\), then \(2012x^{2011}dx = dy\). We can see that our integral now becomes: \[v=\int\frac{1}{y^2+1}dy\] This integral is the derivative of the inverse tangent function, so we get: \[v = \tan^{-1}(y) = \tan^{-1}(x^{2012})\]
03

Apply the integration by parts formula

Now we can apply the integration by parts formula: \[\int \frac{x^{2012} \tan^{-1}(x^{2012})}{1+x^{4024}} dx = uv - \int v du\] \[\int \frac{x^{2012} \tan^{-1}(x^{2012})}{1+x^{4024}} dx = (\tan^{-1}(x^{2012}))(\tan^{-1}(x^{2012})) - \int (\tan^{-1}(x^{2012}))(\frac{2012x^{2011}}{1+x^{4024}}dx)\] Notice that we get back the same term we started with on the right side: \[\int \frac{x^{2012} \tan^{-1}(x^{2012})}{1+x^{4024}} dx = (\tan^{-1}(x^{2012}))(\tan^{-1}(x^{2012})) - k\times (\int \frac{x^{2012} \tan^{-1}(x^{2012})}{1+x^{4024}} dx)\] In order to match with the given formula, the value of k should be \(\frac{1}{2012}\). Therefore, the correct answer is (a) \([1/(2012)]\).

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Most popular questions from this chapter

\(\int\left[(x+1) d x /\left\\{x\left(1+x e^{x}\right)^{2}\right\\}\right]=\) (a) \(\log \left(\mathrm{xe}^{\mathrm{x}}\right) /\left(1+\mathrm{xe}^{\mathrm{x}}\right) \mid-\left[1 /\left(1+\mathrm{xe}^{\mathrm{x}}\right)\right]\) (b) \(\log \left|\left(x e^{x}\right) /\left(1+x e^{x}\right)\right|-\left[1 /\left(1+x e^{x}\right)\right]\) (c) \(\log \left(\mathrm{xe}^{\mathrm{x}}\right) /\left(1+\mathrm{xe}^{\mathrm{x}}\right) \mid+\left[1 /\left(1+\mathrm{xe}^{\mathrm{x}}\right)\right]\) (d) \(\log \left|\left(1+x e^{x}\right) /\left(x e^{x}\right)\right|-\left[1 /\left(1+x e^{x}\right)\right]\)

\(\int[\log (\log x)+\\{1 /(\log x)\\}] d x=\) \(+c\) (a) \([\mathrm{x} /\\{\log (\log \mathrm{x})\\}]\) (b) \(\mathrm{x}+\log (\log \mathrm{x})\) (c) \(\log (\log \mathrm{x})+(1 / \mathrm{x})\) (d) \(x \log (\log x)\)

\(\int \mathrm{e}^{\mathrm{x}}\left[(1-\mathrm{x}) /\left(1+\mathrm{x}^{2}\right)\right]^{2} \mathrm{dx}=\underline{\mathrm{c}}\) (a) \(e^{x}\left(1+x^{2}\right)\) (b) \(\left[\mathrm{e}^{\mathrm{x}} /\left(1+\mathrm{x}^{2}\right)\right]\) (c) \(\mathrm{e}^{\mathrm{x}}\left[(1-\mathrm{x}) /\left(1+\mathrm{x}^{2}\right)\right]\) (d) \(e^{x}\left(1-x^{2}\right)\)

\(I=\int\left[\left(x^{3} d x\right) /\left\\{\sqrt{\left. \left.\left(1+x^{8}\right)\right\\}\right]}\right.\right.\) (a) \(\log \mid x^{4}+\sqrt{\left(1+x^{8}\right) \mid+c}\) (b) \(\log \mid \sqrt{\left(x^{8}+1\right) \mid+c}\) (c) \((1 / 4) \log \mid \mathrm{x}^{4}+\sqrt{\left(1+\mathrm{x}^{8}\right) \mid+\mathrm{c}}\) (d) none of these

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