What is the integral from-1 to 1 of 4/(1+x^2) ?

The integral from-1 to 1 of 4/(1+x^2) is 2pi

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integral from-1 to 1 of 4/(1+x^2)

\int_{- 1}^{1} \frac{4}{1 + x ^{2}} d x

2 π

Decimal

6.28318 \dots

Solution steps

\int_{- 1}^{1} \frac{4}{1 + x ^{2}} d x

Take the constant out:

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

= 4 \cdot \int_{- 1}^{1} \frac{1}{1 + x ^{2}} d x

Use the common integral:

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

= 4 [arctan (x)]_{- 1}^{1}

Compute the boundaries:

\frac{π}{2}

= 4 \cdot \frac{π}{2}

Simplify

= 2 π

f (x) = x^{6} e^{x}

sin (x) ln (8 x)

\frac{d}{d x} (e^{x} + 4)

\int_{- 3}^{3} (9 - x^{2})^{2} d x

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

What is the integral from-1 to 1 of 4/(1+x^2) ?
The integral from-1 to 1 of 4/(1+x^2) is 2pi