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

The integral from 1 to 3 of t/(sqrt(3+t^2)) is 2(sqrt(3)-1)

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

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

2 (3 - 1)

Decimal

1.46410 \dots

Solution steps

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

Apply u-substitution

= \int_{4}^{12} \frac{1}{2 u} d u

Take the constant out:

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

= \frac{1}{2} \cdot \int_{4}^{12} \frac{1}{u} d u

Apply the Power Rule

= \frac{1}{2} [2 u^{\frac{1}{2}}]_{4}^{12}

Apply radical rule:

= \frac{1}{2} [2 u]_{4}^{12}

Compute the boundaries:

43 - 4

= \frac{1}{2} (43 - 4)

Simplify

= 2 (3 - 1)

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