How do you integrate $\int \frac{1}{x \sqrt{3 + x^{2}}} d x$ $int 1/(xsqrt(3 + x^2))dx$ using trigonometric substitution?

Question

How do you integrate $\int \frac{1}{x \sqrt{3 + x^{2}}} d x$ $int 1/(xsqrt(3 + x^2))dx$ using trigonometric substitution?

1 Answer

Impact of this question

2325 views around the world

You can reuse this answer
Creative Commons License

Answer 1 · 2018-03-18T07:23:35

The answer is $= ln (∣ ∣ ∣ \frac{x}{\sqrt{3 + x^{2}} + \sqrt{3}} ∣ ∣ ∣) + C$ $=ln(|x/(sqrt(3+x^2)+sqrt3)|)+C$

Explanation:

The denominator is

$x \sqrt{3 + x^{2}} = x \sqrt{3 (1 + {(\frac{x}{\sqrt{3}})}^{2})}$ $xsqrt(3+x^2)=xsqrt(3(1+(x/sqrt3)^2))$

$= \sqrt{3} x \sqrt{(1 + {(\frac{x}{\sqrt{3}})}^{2})}$ $=sqrt3xsqrt((1+(x/sqrt3)^2))$

Perform the substitution

$\frac{x}{\sqrt{3}} = tan θ$ $x/sqrt3=tantheta$

$\frac{d x}{\sqrt{3}} = {sec}^{2} θ d θ$ $dx/sqrt3=sec^2theta d theta$

$\sqrt{(1 + {(\frac{x}{\sqrt{3}})}^{2})} = \sqrt{1 + {tan}^{2} θ} = \sqrt{{sec}^{2} θ} = sec θ$ $sqrt((1+(x/sqrt3)^2))=sqrt(1+tan^2theta)=sqrt(sec^2theta)=sectheta$

Therefore, the integral is

$I = \int \frac{d x}{x \sqrt{3 + x^{2}}} = \int \frac{\sqrt{3} {sec}^{2} θ d θ}{\sqrt{3} tan θ sec θ}$ $I=int(dx)/(xsqrt(3+x^2))=int(sqrt3sec^2theta d theta)/(sqrt3 tan thetasectheta)$

$= \int \frac{sec θ d θ}{tan θ}$ $=int(secthetad theta)/tan theta$

$= \int \frac{d θ}{sin θ}$ $=int(d theta)/sin theta$

$= \int (csc θ d θ)$ $=int(csctheta d theta)$

$= \int \frac{csc θ (csc θ + cot θ) d θ}{csc θ + cot θ}$ $=int(csctheta(csctheta+cottheta)d theta)/(csctheta+cottheta)$

$= \int \frac{({csc}^{2} θ + csc θ cot θ) d θ}{csc θ + cot θ}$ $=int((csc^2theta+csc thetacottheta)d theta)/(csctheta+cottheta)$

Perform the substitution

$v = csc θ + cot θ$ $v=csctheta+cottheta$ , $\Rightarrow$ $=>$ , $d v = (- cot θ csc θ - {csc}^{2} θ) d θ$ $dv=(-cotthetacsctheta-csc^2theta)d theta$

Therefore,

$I = \int \frac{- d v}{v}$ $I=int(-dv)/(v)$

$= - ln (v)$ $=-ln(v)$

$= - ln (csc θ + cot θ)$ $=-ln(csctheta+cottheta)$

$tan θ = \frac{x}{\sqrt{3}}$ $tan theta=x/sqrt3$ , $\Rightarrow$ $=>$ , $csc θ = \frac{\sqrt{3 + x^{2}}}{x}$ $csctheta=sqrt(3+x^2)/x$

$tan θ = \frac{x}{\sqrt{3}}$ $tan theta=x/sqrt3$ , $\Rightarrow$ $=>$ , $cot θ = \frac{\sqrt{3}}{x}$ $cottheta=sqrt(3)/x$

Finally,

$I = - ln (\frac{\sqrt{3 + x^{2}}}{x} + \frac{\sqrt{3}}{x})$ $I=-ln(sqrt(3+x^2)/x+sqrt(3)/x)$

$= - ln (\frac{\sqrt{3 + x^{2}} + \sqrt{3}}{x})$ $=-ln((sqrt(3+x^2)+sqrt3)/(x))$

$= ln (∣ ∣ ∣ \frac{x}{\sqrt{3 + x^{2}} + \sqrt{3}} ∣ ∣ ∣) + C$ $=ln(|x/(sqrt(3+x^2)+sqrt3)|)+C$

Science

Math

Humanities

... and beyond

How do you integrate $\int \frac{1}{x \sqrt{3 + x^{2}}} d x$ $int 1/(xsqrt(3 + x^2))dx$ using trigonometric substitution?

1 Answer

Explanation:

Related questions

Impact of this question

Science

Math

Humanities

... and beyond

How do you integrate ∫1x√3+x2dxint 1/(xsqrt(3 + x^2))dx using trigonometric substitution?

1 Answer

Explanation:

Related questions

Impact of this question

How do you integrate $\int \frac{1}{x \sqrt{3 + x^{2}}} d x$ $int 1/(xsqrt(3 + x^2))dx$ using trigonometric substitution?