What is the integral from 0 to 1 of 3/(1+x^2) ?

The integral from 0 to 1 of 3/(1+x^2) is (3pi)/4

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integral from 0 to 1 of 3/(1+x^2)

\int_{0}^{1} \frac{3}{1 + x ^{2}} d x

\frac{3 π}{4}

Decimal Notation

2.35619 \dots

Solution steps

\int_{0}^{1} \frac{3}{1 + x ^{2}} d x

Take the constant out:

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

= 3 \cdot \int_{0}^{1} \frac{1}{1 + x ^{2}} d x

Use the common integral:

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

= 3 [arctan (x)]_{0}^{1}

Compute the boundaries:

\frac{π}{4}

= 3 \cdot \frac{π}{4}

Simplify

= \frac{3 π}{4}

\int_{0}^{1} x (x^{2} + 1)^{3} d x

\int ln (e x) d x

x \to 1 lim (x^{2} - x + 1)

\int (x + 2)^{4} d x

\frac{d}{d x} ((x^{5} - 3 x)^{2})

What is the integral from 0 to 1 of 3/(1+x^2) ?
The integral from 0 to 1 of 3/(1+x^2) is (3pi)/4