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Chapter 11: Problem 896

\(\int \mathrm{e}^{4 \log \mathrm{x}}\left(\mathrm{x}^{5}+1\right)^{-1}=\underline{\mathrm{c}}\) (a) \((1 / 5) \log \left(\mathrm{x}^{4}+1\right)\) (b) \(-\log \left(\mathrm{x}^{4}+1\right)\) (c) \(\log \left(x^{4}+1\right)\) (d) \((1 / 5) \log \left(x^{5}+1\right)\)

Short Answer

Expert verified
(d) \(\frac{1}{5}\log\left(x^5 + 1\right)\)

Step by step solution

01

Simplify the integrand

Apply the logarithm properties of \(\exp\) and \(\log\). Since \(\mathrm{e}^{4\log\mathrm{x}} = \mathrm{x}^4\), we can rewrite the given integral as: \[\int \mathrm{x}^4\left(\mathrm{x}^{5}+1\right)^{-1}\,\mathrm{d}x\]
02

Integrate using substitution

Let us perform a substitution: \[u = x^5 + 1\] \[ \frac{\mathrm{d}u}{\mathrm{d}x} = 5x^4\] \[\mathrm{d}u = 5x^4\mathrm{d}x\] Substituting into the integral: \[\int \frac{1}{5}\cdot\frac{\mathrm{d}u}{u}\]
03

Integrating

Now we can easily integrate the expression: \[\frac{1}{5} \int \frac{\mathrm{d}u}{u}\] The integral of \(1/u \mathrm{d}u\) is \(\log(|u|) + C\), so our final expression will be: \[\frac{1}{5}\log(|u|) + C\]
04

Reverse substitution

Substitute the original variable \(x\) back in: \[\frac{1}{5}\log(x^5 + 1) + C\] Comparing the options given, it matches with option (d): \(\boxed{\text{(d) }\frac{1}{5}\log\left(x^5 + 1\right)}\)

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Most popular questions from this chapter

\(\int\left[\left(\sec ^{2} x-2009\right) /\left(\sin ^{2009} x\right)\right] d x=\) \(-c\) (a) \(\left[(\cot x) /\left(\sin ^{2009} x\right)\right]\) (b) \(\left[(-\cot x) /\left(\sin ^{2009} x\right)\right]\) (c) \(\left[(\tan x) /\left(\sin ^{2009} x\right)\right]\) (d) \(\left[(-\tan \mathrm{x}) /\left(\sin ^{2009} \mathrm{x}\right)\right]\)

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

If \(\int \sin ^{-1}\left[2 x /\left(1+x^{2}\right)\right] d x=f(x)-\log \left(1+x^{2}\right)+c\) then \(f(x)=\) (a) \(\mathrm{xtan}^{-1} \mathrm{x}\) (b) \(-\mathrm{xtan}^{-1} \mathrm{x}\) (c) \(2 \mathrm{xtan}^{-1} \mathrm{x}\) (d) \(-2 \mathrm{xtan}^{-1} \mathrm{x}\)

\(\int e^{x}\left[\left(x^{3}-x-2\right) /\left(x^{2}+1\right)^{2}\right] d x=\) (a) \(e^{x}\left[(2 x-1) /\left(x^{2}+1\right)\right]\) (b) \(e^{x}\left[(x+1) /\left(x^{2}+1\right)\right]\) (c) \(e^{x}\left[(x-1) /\left(x^{2}+1\right)\right]\) (b) \(e^{x}\left[(2 x-2) /\left(x^{2}+1\right)\right]\)

If \(\int[(1+\cos 8 \mathrm{x}) /(\cot 2 \mathrm{x}-\tan 2 \mathrm{x})] \mathrm{dx}=\mathrm{Acos} 8 \mathrm{x}+\mathrm{C}\) then \(\mathrm{A}=\) (a) \(\overline{(1 / 16)}\) (b) \(-(1 / 8)\) (c) \(-(1 / 16)\) (d) \((1 / 8)\)
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