Question

How do you find the integral of `int 3/(2sqrtx(1+x)`?

Answer 1

`3arctan(sqrtx)+C`

Explanation:

`I=int3/(2sqrtx(1+x))dx=3/2intdx/(sqrtx(1+x))`

If the substitution presented by Eric Sia is not immediately apparent, another substitution we can try is `u=sqrtx`. This implies that `du=1/(2sqrtx)dx` which we already have in the integrand.

`I=3int1/(1+x)1/(2sqrtx)dx=3int1/(1+u^2)du`

At this point, you may recognize that this is the arctangent integral. However, since this is in the "Integration by Trigonometric Substitution" section, we can derive the arctangent integral here using a trig substitution. Let `u=tantheta`. Thus `du=sec^2thetad theta`.

`I=3int1/(1+tan^2theta)(sec^2thetad theta)=3intd theta=3theta+C`

Note that we used `1+tan^2theta=sec^2theta`. Also, reverse the substitution of `u=tantheta` to show that `theta=arctan(u)`. Since `u=sqrtx`, we see that `theta=arctan(sqrtx)`.

`I=3arctan(sqrtx)+C`