Vtyjkyuk

Given improper integral $$int limits_0^inftyleft(frac1sqrtx^2+4-frackx+2right)dx , ,$$

there exists $k$ that makes this integral convergent.

Find its integration value.

Choices are $ln 2$, $ln 3$, $ln 4$, and $ln 5$.

I've written every information from the problem.

Yet I'm not sure whether I should find the integration value from the given integral or $k$.

What I've tried so far is,

$int_0^infty frac1sqrtx^2+4 , dx= left[sinh^-1fracx2right]_0^infty$

How should I proceed?

We have that for $xto infty$

$$frac1sqrtx^2+4=frac1x(1+4/x^2)^-1/2sim frac1x-frac2x^3$$

and

$$frac k x+2sim frac k x$$

therefore in order to have convergence we need $k=1$ in such way that the $frac1x$ term cancels out.

Then we need to solve and evaluate

$$int_0^inftyleft (frac1sqrtx^2+4-frac1x+2right)dx=left[sinh^-1frac x 2 -log (x+2)right]_0^infty$$

and to evaluate the value at $infty$ recall that

$$sinh^-1frac x 2=log left(frac x 2 + sqrtfracx^24+1right) sim log x$$

As shown in gimusi's answer, you do not need to compute the integral.

However, if you want to integrate, consider
$$I=int_0^pleft(frac1sqrtx^2+4-frackx+2right),dx=sinh ^-1left(fracp2right)-k log (p+2)+k log (2)$$ and use series expansion for large $p$. This should give
$$I=(1-k) log left(pright)+k log (2)-frac2 kp+frac2
k+1p^2+Oleft(frac1p^3right)$$ and look at the limit when $pto infty$.

There are no integration issue in a right neighbourhood of the origin, but when $xto +infty$ we have that the integrand function behaves like $frac1-kx+Oleft(frac1x^2right)$, so a necessary and sufficient condition for the integrability is $k=1$. In such a case
$$begineqnarray* int_0^+inftyleft[frac1sqrtx^2+4-frac1x+2right],dx &stackrelxmapsto 2z=& int_0^+inftyleft[frac1sqrtz^2+1-frac1z+1right],dz\[0.3cm]&=&left[textarcsinh(z)-log(z+1)right]_0^+infty\[0.3cm]&=&lim_zto +inftytextarcsinh(z)-log(z+1)\&stackrelzmapstosinh t=&lim_tto +infty logleft(frace^tsinh t+1right)=colorredlog 2.endeqnarray*$$