What is the integral from 0 to 1 of 5/(1+t^2) ?

The integral from 0 to 1 of 5/(1+t^2) is (5pi)/4

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integral from 0 to 1 of 5/(1+t^2)

\int_{0}^{1} \frac{5}{1 + t ^{2}} d t

\frac{5 π}{4}

Decimal Notation

3.92699 \dots

Solution steps

\int_{0}^{1} \frac{5}{1 + t ^{2}} d t

Take the constant out:

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

= 5 \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)

= 5 [arctan (t)]_{0}^{1}

Compute the boundaries:

\frac{π}{4}

= 5 \cdot \frac{π}{4}

Simplify

= \frac{5 π}{4}

\int x^{5} (x^{6} + 5)^{5} d x

ln (x^{2} - 1)

f (x) = x^{2} e^{11 x}

\int \frac{x ^{3}}{x ^{4} + 1} d x

x \to 0 lim (\frac{tan ( x )}{5 x})

What is the integral from 0 to 1 of 5/(1+t^2) ?
The integral from 0 to 1 of 5/(1+t^2) is (5pi)/4