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}

Decimal Notation

2.35619 \dots

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}

3 x cos (x)

- 4 - 9 x^{2}

2 \frac{d ^{2} y}{d x ^{2}} + 11 \frac{d y}{d x} - 6 y = 8 x

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

x \to 2 lim (x^{3} + 5 x^{2} - 7 x + 1)

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