# Evaluate the integral integral square root 1 + x 1 - x d x

## Question:

Evaluate the integral {eq}\int \sqrt{\frac{1+x}{1-x}} \ dx {/eq}

## Integration:

Integration is the reverse of differentiation. Antiderivatives are related to definite integrals through the fundamental theorem of calculus: the definite integral of a function over an interval is equal to the difference between the values of an antiderivative evaluated at the endpoints of the interval. The discrete equivalent of the notion of antiderivative is antidifference.

## Answer and Explanation:

{eq}\int \sqrt{\frac{1+x}{1-x}} \ dx = \int \frac{\sqrt{1+x}}{\sqrt{1-x}} * \frac{\sqrt{1+x}}{\sqrt{1+x}} \ dx \\ =\int \frac{1+x}{\sqrt{1-x^2}}\ dx \\ =\int \frac{1}{\sqrt{1-x^2}} + \frac{x}{\sqrt{1-x^2}}\ dx \\ = \sin^{-1} x - \frac12 \int (1-x^2)^{-\frac12} d(1-x^2) + C \\ = \sin^{-1} x - \frac12 \int (1-x^2)^{-\frac12} d(1-x^2) + C\\ = \sin^{-1} x - \frac12 (\frac{(1-x^2)^{\frac12}}{\frac12}) +c \\ =\sin^{-1} x - \sqrt{1-x^2} +c \\ {/eq}

#### Learn more about this topic: Integration Problems in Calculus: Solutions & Examples

from AP Calculus AB & BC: Homework Help Resource

Chapter 13 / Lesson 13
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