What is the integral from 0 to 1 of (10)/(x^2+1) ?

The integral from 0 to 1 of (10)/(x^2+1) is (5pi)/2

English

Español

Português

Français

Deutsch

Italiano

Русский

中文(简体)

한국어

日本語

Tiếng Việt

עברית

العربية

Popular >

integral from 0 to 1 of (10)/(x^2+1)

\int_{0}^{1} \frac{10}{x ^{2} + 1} d x

\frac{5 π}{2}

Solution steps

\int_{0}^{1} \frac{10}{x ^{2} + 1} d x

Take the constant out:

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

= 10 \cdot \int_{0}^{1} \frac{1}{x ^{2} + 1} d x

Use the common integral:

\int \frac{1}{x ^{2} + 1} d x = arctan (x)

= 10 [arctan (x)]_{0}^{1}

Compute the boundaries:

\frac{π}{4}

= 10 \cdot \frac{π}{4}

Simplify

= \frac{5 π}{2}

\int_{0}^{2 π} cos^{2} (y) d y

\int_{1}^{9} \frac{4}{x} d x

\int_{1}^{6} \frac{4}{x ^{2}} d x

\int_{2}^{e} \frac{x ^{4} - 3 x}{x ^{2}} d x

\int_{- 1}^{1} x^{3} sin (x) d x

What is the integral from 0 to 1 of (10)/(x^2+1) ?
The integral from 0 to 1 of (10)/(x^2+1) is (5pi)/2