What is the integral from 0 to 12 of 5/(1+0.2t^2) ?

The integral from 0 to 12 of 5/(1+0.2t^2) is 15.50232…

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integral from 0 to 12 of 5/(1+0.2t^2)

\int_{0}^{12} \frac{5}{1 + 0.2 t ^{2}} d t

15.50232 \dots

Solution steps

\int_{0}^{12} \frac{5}{1 + 0.2 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}^{12} \frac{1}{1 + 0.2 t ^{2}} d t

Apply Integral Substitution

= 5 \cdot \int_{0}^{5.36656 \dots} \frac{2.23606 \dots}{1 + u ^{2}} d u

Take the constant out:

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

= 5 \cdot 2.23606 \dots \cdot \int_{0}^{5.36656 \dots} \frac{1}{1 + u ^{2}} d u

Use the common integral:

\int \frac{1}{1 + u ^{2}} d u = arctan (u)

= 5 \cdot 2.23606 \dots [arctan (u)]_{0}^{5.36656 \dots}

Simplify

= 11.18033 \dots [arctan (u)]_{0}^{5.36656 \dots}

Compute the boundaries:

1.38657 \dots

= 11.18033 \dots \cdot 1.38657 \dots

Simplify

= 15.50232 \dots

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What is the integral from 0 to 12 of 5/(1+0.2t^2) ?
The integral from 0 to 12 of 5/(1+0.2t^2) is 15.50232…