Can anyone help me integrate : \int\limits { (x*(2 x^2 + 6 x + 1))/(√(x^2 + 4 x + 1)) } \, dx

Question

Can anyone help me integrate :


`\int\limits { (x*(2 x^2 + 6 x + 1))/(√(x^2 + 4 x + 1)) } \, dx`

Answer 1

Rewrite the second factor in the numerator as


`2x^2+6x+1=2(x+2)^2-2(x+2)-3`

Then in the entire integrand, set
`x+2=\sqrt3\sec t`, so that
`\mathrm dx=\sqrt3\sec t\tan t\,\mathrm dt`. The integral is then equivalent to


`\displaystyle\int((\sqrt3\sec t-2)(6\sec^2t-2\sqrt3\sec t-3))/(√((\sqrt3\sec t)^2-3))(\sqrt3\sec t)\,\mathrm dt`

`=\displaystyle\int((6\sqrt3\sec^3t-18\sec^2t+\sqrt3\sec t+6)\sec t)/(√(\sec^2t-1))\,\mathrm dt`

`=\displaystyle\int((6\sqrt3\sec^3t-18\sec^2t+\sqrt3\sec t+6)\sec t)/(√(\tan^2t))\,\mathrm dt`

`=\displaystyle\int((6\sqrt3\sec^3t-18\sec^2t+\sqrt3\sec t+6)\sec t)/(|\tan t|)\,\mathrm dt`

Note that by letting
`x+2=\sqrt3\sec t`, we are enforcing an invertible substitution which would make it so that
`t=\mathrm{arcsec}(x+2)/(\sqrt3)` requires
`0\le t<\frac\pi2` or
`\frac\pi2<t\le\pi`. However,
`\tan t` is positive over this first interval and negative over the second, so we can't ignore the absolute value.

So let's just assume the integral is being taken over a domain on which
`\tan t>0` so that
`|\tan t|=\tan t`. This allows us to write


`=\displaystyle\int((6\sqrt3\sec^3t-18\sec^2t+\sqrt3\sec t+6)\sec t)/(\tan t)\,\mathrm dt`

`=\displaystyle\int(6\sqrt3\sec^3t-18\sec^2t+\sqrt3\sec t+6)\csc t\,\mathrm dt`

We can show pretty easily that


`\displaystyle\int\csc t\,\mathrm dt=-\ln|\csc t+\cot t|+C`

`\displaystyle\int\sec t\csc t\,\mathrm dt=-\ln|\csc2t+\cot2t|+C`

`\displaystyle\int\sec^2t\csc t\,\mathrm dt=\sec t-\ln|\csc t+\cot t|+C`

`\displaystyle\int\sec^3t\csc t\,\mathrm dt=\frac12\sec^2t+\ln|\tan t|+C`

which means the integral above becomes


`=3\sqrt3\sec^2t+6\sqrt3\ln|\tan t|-18\sec t+18\ln|\csc t+\cot t|-\sqrt3\ln|\csc2t+\cot2t|-6\ln|\csc t+\cot t|+C`

`=3\sqrt3\sec^2t-18\sec t+6\sqrt3\ln|\tan t|+12\ln|\csc t+\cot t|-\sqrt3\ln|\csc2t+\cot2t|+C`

Back-substituting to get this in terms of
`x` is a bit of a nightmare, but you'll find that, since
`t=\mathrm{arcsec}(x+2)/(\sqrt3)`, we get


`\sec t=(x+2)/(\sqrt3)`

`\sec^2t=\frac{(x+2)^2}3`

`\tan t=\sqrt{\frac{x^2+4x+1}3}`

`\cot t=\sqrt{\frac3{x^2+4x+1}}`

`\csc t=(x+2)/(√(x^2+4x+1))`

`\csc2t=((x+2)^2)/(2\sqrt3√(x^2+4x+1))`

etc.