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

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

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

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

\frac{3 π}{4}

Solution steps

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

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 + t ^{2}} d t

Use the common integral:

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

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

Compute the boundaries:

\frac{π}{4}

= 3 \cdot \frac{π}{4}

Simplify

= \frac{3 π}{4}

\frac{d}{d θ} (3 - sin (θ))

f (x) = x^{2} - 4, at x = 0

\frac{d}{d x} (\frac{3 x ^{4} + 1}{x ^{3}})

\frac{d}{d x} (2 - \frac{x ^{2} - 2 x + 4}{x - 2})

\int \frac{ln ( x )}{x} d x

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