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

= \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

\int \frac{4}{( 4 + x ^{2} ) ^{2}} d x

f (x) = x^{2} + 7, (- 4, 23)

x \to 1 + lim (x^{\frac{4}{1 - x}})

\int \frac{- 2 x - 6}{x ^{3} - 2 x ^{2} - 3 x} d x

x \to 1 + lim (\frac{4}{x - 1})

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