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

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

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integral from 2 to 3 of (4t)/(1+t^2)

\int_{2}^{3} \frac{4 t}{1 + t ^{2}} d t

2 ln (2)

Decimal Notation

1.38629 \dots

Solution steps

\int_{2}^{3} \frac{4 t}{1 + t ^{2}} d t

Take the constant out:

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

= 4 \cdot \int_{2}^{3} \frac{t}{1 + t ^{2}} d t

Apply u-substitution

: \int_{5}^{10} \frac{1}{2 u} d u

= 4 \cdot \int_{5}^{10} \frac{1}{2 u} d u

Take the constant out:

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

= 4 \cdot \frac{1}{2} \cdot \int_{5}^{10} \frac{1}{u} d u

Use the common integral:

\int \frac{1}{u} d u = ln (∣ u ∣)

= 4 \cdot \frac{1}{2} [ln ∣ u ∣]_{5}^{10}

Simplify

4 \cdot \frac{1}{2} [ln ∣ u ∣]_{5}^{10} : 2 [ln ∣ u ∣]_{5}^{10}

= 2 [ln ∣ u ∣]_{5}^{10}

Compute the boundaries:

ln (10) - ln (5)

= 2 (ln (10) - ln (5))

Simplify

= 2 ln (2)

\int_{0}^{\frac{π}{2}} sin (x) cos (y) d y

\int_{2}^{x^{2}} t^{2} t^{2} + 9 d t

\int_{0}^{10 π} ∣ sin (x) ∣ d x

\int_{0}^{2} sin (\frac{nπ}{2} x) d x

\int_{0}^{\frac{5}{3}} 3 x^{2} - 5 x d x

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