What is the integration of $\frac{d x}{x . \sqrt{x^{3} + 4}}$ $(dx)/(x.sqrt(x^3+4))$ ??

Question

What is the integration of $\frac{d x}{x . \sqrt{x^{3} + 4}}$ $(dx)/(x.sqrt(x^3+4))$ ??

1 Answer

Impact of this question

2766 views around the world

You can reuse this answer
Creative Commons License

Answer 1 · 2018-03-07T00:35:15

$\frac{1}{6} ln ∣ ∣ ∣ \frac{\sqrt{x^{3} + 4} - 2}{\sqrt{x^{3} + 4} + 2} ∣ ∣ ∣ + C$ $1/6 ln|{sqrt(x^3+4)-2}/{sqrt(x^3+4)+2}|+C$

Explanation:

Substitute $x^{3} + 4 = u^{2}$ $x^3+4=u^2$ . Then $3 x^{2} d x = 2 u d u$ $3x^2dx=2udu$ , so that

$\frac{d x}{x \sqrt{x^{3} + 4}} = \frac{2 u d u}{3 x^{3} u} = \frac{2}{3} \frac{d u}{u^{2} - 4} = \frac{1}{6} (\frac{d u}{u - 2} - \frac{d u}{u + 2})$ $dx/{x sqrt{x^3+4}} = {2udu}/{3x^3u} = 2/3 {du}/(u^2-4) = 1/6({du}/{u-2}-{du}/{u+2})$

Thus

$\int \frac{d x}{x \sqrt{x^{3} + 4}} = \frac{1}{6} \int (\frac{d u}{u - 2} - \frac{d u}{u + 2}) = \frac{1}{6} ln ∣ ∣ ∣ \frac{u - 2}{u + 2} ∣ ∣ ∣ + C$ $int dx/{x sqrt{x^3+4}} = 1/6 int ({du}/{u-2}-{du}/{u+2})=1/6 ln|{u-2}/{u+2}|+C$
$= \frac{1}{6} ln ∣ ∣ ∣ \frac{\sqrt{x^{3} + 4} - 2}{\sqrt{x^{3} + 4} + 2} ∣ ∣ ∣ + C$ $=1/6 ln|{sqrt(x^3+4)-2}/{sqrt(x^3+4)+2}|+C$

Science

Math

Humanities

... and beyond

What is the integration of $\frac{d x}{x . \sqrt{x^{3} + 4}}$ $(dx)/(x.sqrt(x^3+4))$ ??

1 Answer

Explanation:

Related questions

Impact of this question

Science

Math

Humanities

... and beyond

What is the integration of dxx.√x3+4(dx)/(x.sqrt(x^3+4))??

1 Answer

Explanation:

Related questions

Impact of this question

What is the integration of $\frac{d x}{x . \sqrt{x^{3} + 4}}$ $(dx)/(x.sqrt(x^3+4))$ ??