What is the integral from 0 to 1 of ((9t)/(1+t^2)) ?

The integral from 0 to 1 of ((9t)/(1+t^2)) is 9/2 ln(2)

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integral from 0 to 1 of ((9t)/(1+t^2))

\int_{0}^{1} (\frac{9 t}{1 + t ^{2}}) d t

\frac{9}{2} ln (2)

Decimal Notation

3.11916 \dots

Solution steps

\int_{0}^{1} \frac{9 t}{1 + t ^{2}} d t

Take the constant out:

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

= 9 \cdot \int_{0}^{1} \frac{t}{1 + t ^{2}} d t

Apply u-substitution

: \int_{1}^{2} \frac{1}{2 u} d u

= 9 \cdot \int_{1}^{2} \frac{1}{2 u} d u

Take the constant out:

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

= 9 \cdot \frac{1}{2} \cdot \int_{1}^{2} \frac{1}{u} d u

Use the common integral:

\int \frac{1}{u} d u = ln (∣ u ∣)

= 9 \cdot \frac{1}{2} [ln ∣ u ∣]_{1}^{2}

Simplify

9 \cdot \frac{1}{2} [ln ∣ u ∣]_{1}^{2} : \frac{9}{2} [ln ∣ u ∣]_{1}^{2}

= \frac{9}{2} [ln ∣ u ∣]_{1}^{2}

Compute the boundaries:

ln (2)

= \frac{9}{2} ln (2)

y^{^{'}} = x (3 - y)

\int (x^{3} (x^{2} - 2 x)^{3}) d x

y = \frac{1}{2} (e^{x} - e^{- x})

925 + 20 x - 0.079 x^{2}

\int_{- 9}^{0} \frac{1}{16 - x} d x

What is the integral from 0 to 1 of ((9t)/(1+t^2)) ?
The integral from 0 to 1 of ((9t)/(1+t^2)) is 9/2 ln(2)