Question

How can i solve this integral? `int (2/(5x)sqrt (3+4/x^2)dx`

Answer 1

I got: `(2sqrt3)/5lnabs((3x^2+4)/(2sqrt3x)+(sqrt3x)/2-2/(sqrt3x))+C`

Explanation:

`I=int2/(5x)sqrt(3+4/x^2)dx`

Use the substitution `x=2/sqrt3cottheta`. This implies that `dx=(-2)/sqrt3cotthetacscthetad theta`.

`I=2/5intsqrt3/2tanthetasqrt(3+4(sqrt3/2tantheta)^2)((-2)/sqrt3cotthetacscthetad theta)`

Canceling `sqrt3/2tantheta(2/sqrt3cottheta)=1`:

`I=(-2)/5intcscthetasqrt(3+3tan^2theta)d theta`

Factoring `sqrt3`, the radical becomes `sqrt(1+tan^2theta)=sectheta`:

`I=(-2sqrt3)/5int(d theta)/(sinthetacostheta)`

Note that `sinthetacostheta=1/2sin2theta`:

`I=(-4sqrt3)/5int(d theta)/(sin2theta)`

`I=(-4sqrt3)/5intcsc2thetad theta`

The integral of cosecant is fairly standard. Look it up here if you aren't sure.

Note that I'll use the form `intcscx=-lnabs(cscx+cotx)`, which is equivalent to `lnabs(cscx-cotx)`. I'm choosing the first one so the minus sign will cancel with the minus sign already outside the integral.

You'll also need to undo the `2theta` by multiplying the integral by `1//2`.

`I=(2sqrt3)/5lnabs(csc2theta+cot2theta)+C`

Use `csc2theta=1/(sin2theta)=1/(2sinthetacostheta)` and `tan2theta=(2tantheta)/(1-tan^2theta)`:

`I=(2sqrt3)/5lnabs(1/(2sinthetacostheta)+(1-tan^2theta)/(2tantheta))+C`

Before proceeding we can factor `2` from both these denominators and remove `(-2sqrt3)/5ln2` from the logarithm using `log(A//B)=logA-logB`. This constant is absorbed into `C`.

`I=(2sqrt3)/5lnabs(cscthetasectheta+cottheta-tantheta)+C`

Our original substitution was `cottheta=(sqrt3x)/2`. Thus:

• `tantheta=1/cottheta=2/(sqrt3x)`
• `csctheta=sqrt(cot^2theta+1)=sqrt(3x^2+4)/2`
• `sectheta=sqrt(tan^2theta+1)=sqrt(3x^2+4)/(sqrt3x)`

So:

`I=(2sqrt3)/5lnabs((3x^2+4)/(2sqrt3x)+(sqrt3x)/2-2/(sqrt3x))+C`