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The question is asking me to do $\int_{0}^{\infty}{4x\over(1+x)^5}dx$

My question is, is there any way to do this without partial fractions. If there is a formula for equations of this type, I will gladly memorize it, I just don't think I'll have time to partial fraction expand a question like this on the P exam.

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    You could try the obvious substitution $y=1+x$ and see if that helps. – 2017-02-28
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    Let $u=1+x$.$ $ – 2017-02-28
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    Oh... wait... then x = u-1...? – 2017-02-28
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    That's correct. – 2017-02-28
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    @BobbieD, I may have over thought the problem, thank you! – 2017-02-28
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    Not a problem.$ $ – 2017-02-28

4 Answers 4

3

Let $u=1+x,$ then $du=dx$ and so your integral is now $$\int_{u=1}^{\infty}4(u-1)u^{-5}du=4\int_{1}^{\infty}(u^{-4}-u^{-5})du=\frac{1}{3}$$

2

For any $a>0$ $$ \int_{0}^{+\infty}\frac{4x}{(a+x)^3}\,dx = \frac{2}{a}\tag{1} $$ hence by applying $\frac{d^2}{da^2}$ to both sides of $(1)$ $$ 12\int_{0}^{+\infty}\frac{4x}{(a+x)^5}\,dx = \frac{4}{a^3}\tag{2} $$ and by evaluating at $a=1$ $$ \int_{0}^{+\infty}\frac{4x\,dx}{(1+x)^5}=\color{red}{\frac{1}{3}}.\tag{3}$$

2

Integrating by parts, $$ \int_0^{\infty} \frac{4x}{(1+x)^5} \, dx = \left[ -\frac{x}{(1+x)^4} \right]_0^{\infty} + \int_0^{\infty} \frac{dx}{(1+x)^4} \\ = 0 + \left[ -\frac{1}{3(1+x)^3} \right]_0^{\infty} = \frac{1}{3} $$

1

Overkill: One finds by integrating by parts that $$ \frac{1}{(1+x)^5} = \frac{1}{4!}\int_0^{\infty} s^4 e^{-s(1+x)} \, ds. $$ We may insert this and change the order of integration to find $$ I = \frac{1}{4!}\int_0^{\infty} s^4 e^{-s} \int_0^{\infty} 4xe^{-sx} \, dx \, ds, $$ integrate by parts again to find $\int_0^{\infty} xe^{-sx} \, dx = s^{-2}$, and then integrate by parts a few more times to conclude that $$ I = \frac{1}{3!}\int_0^{\infty} s^2 e^{-s} \, ds = \frac{2!}{3!} = \frac{1}{3}. $$