What is #int (8x)/sqrt(16+x^2) dx#?

1 Answer
Jun 15, 2016

#8sqrt(16+x^2)+C.#

Explanation:

Let #I=int(8x)/sqrt(16+x^2)dx.#

We use Substitution #: 16+x^2=t^2.#

#:. 2xdx=2tdt.#, or, #xdx=tdt.#

Therefore, #I=int(8x)/sqrt(16+x^2)dx,=int(8t)/sqrtt^2dt=8intdt=8t=8sqrt(16+x^2)+C.#

OR

We can use the Formula : #int(f(x))^n*f'(x)dx=(f(x))^(n+1)/(n+1), n!=-1.#

Take #f(x)=(16+x^2),# #n=-1/2!=-1.# Then, #f'(x)=2x.# Thus,
#I=int(8x)/sqrt(16+x^2)dx=4int(16+x^2)^(-1/2)2xdx=4[(16+x^2)^{-1/2+1}]/(-1/2+1)=4{(16+x^2)^(1/2)}/(1/2)=8sqrt(16+x^2)+C,# as before!