How do you find the indefinite intergral of (1/ sq. root (x+1) dx?

Question

I need help figuring out the steps to reach the answer i know that the answer is 2 sq. root (x+1) + C

Answer 1

Use substitution again.
Let u = x+1, then du = dx
The integral becomes
1/sqrt(u) du
= 1/u^(1/2) du
= u^(-1/2) du
= u^(-1/2 + 1)/(-1/2 + 1) + C
= u^(1/2)/(1/2) + C
= 2sqrt(u) + C
= 2sqrt(x+1) + C

Answer 2

The easiest way is to look it up in the integral tables. In this case, however, it can be done by inspection. The indefinite integral is the same thing as the anti-derivative. So you just ask yourself. "What function's derivative is 1/sq.root(x+1)?" It's pretty easy to see by applying the simple algebraic rules of taking the derivative that it's the derivative of 2 sq.root(x+1). This is because sq.root(x+1) = (x+1)^(1/2) and when you take its derivative you multiply by the exponent (1/2) and subtract 1 from the exponent. That gives you (1/2)(x+1)^(-1/2), which is 1/[2 sq.root(x+1)]. You also multiply by the derivative of (x+1) with respect to x, but that's just equal to 1 in this case. You need the factor of 2 multiplier to cancel the exponent of 1/2 when you multiply by it.

I'm guessing that I just did a homework problem for you. :)

Answer 3

I=∫1/sq.root(x+1)dx
let √(x+1)=u
by differentiation
1/2*1/√(X+1)dx=du
dx=2√(x+1)du
=2udu
I=∫2udu/u
=∫2du
=2∫du
=2u+c
=2√(x+1)+c