What is the integral of 1/(sqrt(x+1)+x-1) ?

The integral of 1/(sqrt(x+1)+x-1) is 2(1/3 ln|sqrt(x+1)-1|+2/3 ln|sqrt(x+1)+2|)+C

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integral of 1/(sqrt(x+1)+x-1)

\int \frac{1}{x + 1 + x - 1} d x

2 (\frac{1}{3} ln x + 1 - 1 + \frac{2}{3} ln x + 1 + 2) + C

Solution steps

\int \frac{1}{x + 1 + x - 1} d x

Apply u-substitution

= \int \frac{2 u}{u ^{2} + u - 2} d u

Take the constant out:

\int a \cdot f (x) d x = a \cdot \int f (x) d x

= 2 \cdot \int \frac{u}{u ^{2} + u - 2} d u

Take the partial fraction of

\frac{u}{u ^{2} + u - 2} : \frac{1}{3 ( u - 1 )} + \frac{2}{3 ( u + 2 )}

= 2 \cdot \int \frac{1}{3 ( u - 1 )} + \frac{2}{3 ( u + 2 )} d u

Apply the Sum Rule:

\int f (x) \pm g (x) d x = \int f (x) d x \pm \int g (x) d x

= 2 (\int \frac{1}{3 ( u - 1 )} d u + \int \frac{2}{3 ( u + 2 )} d u)

\int \frac{1}{3 ( u - 1 )} d u = \frac{1}{3} ln ∣ u - 1 ∣

\int \frac{2}{3 ( u + 2 )} d u = \frac{2}{3} ln ∣ u + 2 ∣

= 2 (\frac{1}{3} ln ∣ u - 1 ∣ + \frac{2}{3} ln ∣ u + 2 ∣)

Substitute back

u = x + 1

= 2 (\frac{1}{3} ln x + 1 - 1 + \frac{2}{3} ln x + 1 + 2)

Add a constant to the solution

= 2 (\frac{1}{3} ln x + 1 - 1 + \frac{2}{3} ln x + 1 + 2) + C

What is the integral of 1/(sqrt(x+1)+x-1) ?
The integral of 1/(sqrt(x+1)+x-1) is 2(1/3 ln|sqrt(x+1)-1|+2/3 ln|sqrt(x+1)+2|)+C