What is the indefinite integral of {[square root(x+1)] +1}^(-1)?

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giorgiana1976 | College Teacher | (Level 3) Valedictorian

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Let f(x) = 1/[1+sqrt(x+1)]

Int f(x) dx =Int dx/[1+sqrt(x+1)]

We'll replace sqrt(x + 1) by t.

We'll raise to square both sides:

x + 1 = t^2

x = t^2 - 1

We'll differentiate both sides:

dx = 2tdt

We'll re-write the integral in t:

Int 2tdt/(1+t) = 2Int tdt/(1+t)

We'll add and subtract 1 to numerator of the ratio and we'll split the fraction in 2 fractions, using the property of integral to be additive:

2Int (t + 1 - 1)dt/(1+t) = 2Int (t+1)dt/(1+t) - 2Int dt/(t+1)

We'll simplify and we'll get:

Int f(x) dx = 2Int dt - 2ln |t + 1|

We'll change the variable t:

Int f(x) dx = 2*sqrt(x + 1) - 2ln [sqrt(x + 1) + 1] + C

Int f(x) dx = 2*{sqrt(x + 1) - ln [sqrt(x + 1) + 1]}+ C

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