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Evaluate the integral: integral_0^1 -1/x^2+1 dx

Question:

Evaluate the integral:

{eq}\displaystyle \int_0^1 \frac{-1}{x^2+1}\ dx {/eq}

Intgerations

We know that

{eq}\displaystyle \begin{align} \int \frac{1}{x^2+1}dx=\arctan \left(x\right)\\ \end{align} {/eq}

This is a simple integral and We get the solution by using the above formula

After integrations we substitute the given limits

Answer and Explanation:

Given

{eq}\displaystyle \int_0^1 \frac{-1}{x^2+1}\ dx {/eq}

We know that

{eq}\displaystyle \begin{align} \int \frac{1}{x^2+1}dx=\arctan \left(x\right) \end{align} {/eq}

Therefore

{eq}\displaystyle \begin{align} \int_0^1 \frac{-1}{x^2+1}\ dx&=-\left[\arctan \left(x\right)\right]^1_0\\ &=-\left(\arctan(1)-\arctan(0)\right)\\ &=-\frac{\pi }{4} \end{align} {/eq}

Therefore

{eq}\displaystyle \color{blue}{\int_0^1 \frac{-1}{x^2+1}\ dx=\frac{\pi}{4}} {/eq}


Learn more about this topic:

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Integration Problems in Calculus: Solutions & Examples

from AP Calculus AB & BC: Homework Help Resource

Chapter 13 / Lesson 13
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