Resolved: "EvaluateA=∫−1+1x2−2x2+1dx"

Rex Maxwell

Answered question

2022-04-05

Evaluate

A = \int_{- 1}^{+ 1} x^{2} - \frac{2}{x^{2} + 1} d x

Kingston Lowery

Beginner2022-04-06Added 10 answers

Your u-sub is off. If

u = x^{2} + 1

then

d u = 2 x d x

. But there is no

2 x

term in your integral, so you'll need to solve for x using u. You'll get

x = \sqrt{u - 1}

, meaning

2 d x = \frac{d u}{\sqrt{u - 1}}

and

\int_{- 1}^{1} \frac{2}{x^{2} + 1} d x = \int \frac{1}{u \sqrt{u - 1}} d u

This integral in terms of u is also not so easy to integrate.
Instead, I would begin with

\int_{- 1}^{1} \frac{2}{x^{2} + 1} d x

and use trig-sub. Imagine a right triangle with angle

θ

, opposite leg length x and adjacent leg length of 1. Then the hypotenuse of this triangle will have length

\sqrt{x^{2} + 1}

, meaning:

\sec^{2} (θ) = x^{2} + 1

\tan (θ) = x

\sec^{2} (θ) d θ = d x

and plugging all this in yields

\int_{- 1}^{1} \frac{2}{x^{2} + 1} d x = 2 \int \frac{1}{\sec^{2} (θ)} \cdot \sec^{2} (θ) d θ = 2

\int d θ = 2 [θ + C]

from the equation

\tan (θ) = x

we then know

θ = \arctan (x)

, so

2 [θ + C] = 2 [\arctan (x) + C]

yielding the expected result.

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