Dtyjlui

Following function is given in calculus and should be solved by using substitution

$$int frac{sqrt{4x^2-1}}{x^4} dx$$

I tried to substitute each component but still did not get answer

hint

Put

$$2x=cosh(t) text{ in } [frac 12,+infty)$$

or
$$2x=-cosh(t) text{ in } (-infty,-frac 12]$$
to get

$$sqrt{4x^2-1}=sqrt{cosh^2(t)-1}=sinh(t)$$

Hint. $displaystyleintfrac{sqrt{4x^2-1}}{x^4}dx=begin{cases}displaystyleintfrac{dx}{x^3}sqrt{4-frac1{x^2}},&x>0\displaystyle-intfrac{dx}{x^3}sqrt{4-frac1{x^2}},&x<0end{cases}$

Keep $4-frac1{x^2}=timplies dt=2dx/x^3$.

This one is actually pretty neat
$$I=intfrac{sqrt{4x^2-1}}{x^4}mathrm dx$$
The first thing to do is get rid of the square root, so we do
$$x=frac12sec uRightarrow mathrm dx=frac12sec utan u,mathrm du$$
Hence
$$I=frac12intfrac{sqrt{4cdotfrac14sec^2u-1}}{frac1{2^4}sec^4u}sec utan u,mathrm du$$
Which simplifies to
$$I=8intfrac{sqrt{sec^2u-1}}{sec^3u}tan u,mathrm du$$
Noting that $sec^2u-1=tan^2u$, we can simplify more:
$$I=8intfrac{tan^2u}{sec^3u}mathrm du$$
Using the reciprocal and quotient identities this simplifies to
$$I=8intsin^2ucos u,mathrm du$$
Then we preform another substitution
$$w=sin uRightarrow mathrm dw=cos u, mathrm du$$
So we have
$$I=8int w^2mathrm dw$$
$$I=frac83w^3$$
$$I=frac83sin^3u$$
$$I=frac83(1-cos^2u)^{3/2}$$
$$I=frac83bigg(1-frac1{sec^2u}bigg)^{3/2}$$
since $x=frac12sec u$,
$$I=frac83bigg(1-frac1{4x^2}bigg)^{3/2}$$
$$I=frac{(4x^2-1)^{3/2}}{3x^3}+C$$