What is the integral from 5 to 10 of (2t)/(t^2+1) ?

The integral from 5 to 10 of (2t)/(t^2+1) is ln(101)-ln(26)

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integral from 5 to 10 of (2t)/(t^2+1)

\int_{5}^{10} \frac{2 t}{t ^{2} + 1} d t

ln (101) - ln (26)

Solution steps

\int_{5}^{10} \frac{2 t}{t ^{2} + 1} d t

Take the constant out:

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

= 2 \cdot \int_{5}^{10} \frac{t}{t ^{2} + 1} d t

Apply u-substitution

: \int_{26}^{101} \frac{1}{2 u} d u

= 2 \cdot \int_{26}^{101} \frac{1}{2 u} d u

Take the constant out:

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

= 2 \cdot \frac{1}{2} \cdot \int_{26}^{101} \frac{1}{u} d u

Use the common integral:

\int \frac{1}{u} d u = ln (∣ u ∣)

= 2 \cdot \frac{1}{2} [ln ∣ u ∣]_{26}^{101}

Simplify

2 \cdot \frac{1}{2} [ln ∣ u ∣]_{26}^{101} : [ln ∣ u ∣]_{26}^{101}

= [ln ∣ u ∣]_{26}^{101}

Compute the boundaries:

ln (101) - ln (26)

= ln (101) - ln (26)

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What is the integral from 5 to 10 of (2t)/(t^2+1) ?
The integral from 5 to 10 of (2t)/(t^2+1) is ln(101)-ln(26)