Infinite intervals of integration Evaluate the

Chapter 7, Problem 17E

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QUESTION:

Infinite intervals of integration Evaluate the following integrals or state that they diverge.

$\int_{0}^{\infty} \frac{x}{\sqrt{x^{4}+1}} d x$

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QUESTION:

Infinite intervals of integration Evaluate the following integrals or state that they diverge.

$\int_{0}^{\infty} \frac{x}{\sqrt{x^{4}+1}} d x$

ANSWER:

Problem 17E

Infinite intervals of integration
Evaluate the following integrals or state that they diverge.
$\int\limits_{0}^{\infty{}}\frac{x}{\sqrt{{x}^{4}+1}}dx$
Solution:
Step 1
To evaluate the following integral, we use method of integration by substitution.
We substitute $x=\sqrt{tant}$
Squaring both sides, we get
${x}^{2}=tant$
Differentiating ${x}^{2}=tant$ , we get
${2xdx}^{\ }=se{c}^{2}t\ dt$
Arranging I as follows so that we can substitute.
$I=\int\limits_{\ }^{\ }\frac{x}{\sqrt{{x}^{4}+1}}dx$
$=\frac{1}{2}\int\limits_{\ }^{\ }\frac{2x\ dx}{\sqrt{{x}^{4}+1}}$
$=\frac{1}{2}\int\limits_{\ }^{\ }\frac{2x\ dx}{\sqrt{{x}^{4}+1}}$
Substituting these values in the integral $\frac{1}{2}\int\limits_{\ }^{\ }\frac{2x\ dx}{\sqrt{{x}^{4}+1}}$ , we get
$I=\frac{1}{2}\int\limits_{\ }^{\ }\frac{se{c}^{2}t\ dt}{\sqrt{ta{n}^{2}t+1}}$
$=\frac{1}{2}\int\limits_{\ }^{\ }\frac{se{c}^{2}t\ dt}{\sqrt{se{c}^{2}t}}$
$=\frac{1}{2}\int\limits_{\ }^{\ }\frac{se{c}^{2}t\ dt}{sec\ t}$
$=\frac{1}{2}\int\limits_{\ }^{\ }sect\ dt$
$=\frac{1}{2}\left({ln\ \left({tant+sect}\right)}\right)+C$