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

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

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

\int_{1}^{2} \frac{t}{t ^{2} - 1} d t

3

Decimal

1.73205 \dots

Solution steps

\int_{1}^{2} \frac{t}{t ^{2} - 1} d t

Apply u-substitution

: \int_{0}^{3} \frac{1}{2 u} d u

= \int_{0}^{3} \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_{0}^{3} \frac{1}{u} d u

Apply the Power Rule

: [2 u^{\frac{1}{2}}]_{0}^{3}

= \frac{1}{2} [2 u^{\frac{1}{2}}]_{0}^{3}

Apply radical rule:

a^{\frac{1}{n}} = n a

= \frac{1}{2} [2 u]_{0}^{3}

Compute the boundaries:

23

= \frac{1}{2} \cdot 23

Simplify

= 3

\int_{1}^{2} 1

\int_{- 2}^{2} 1 - \frac{x ^{2}}{4} d x

\int_{1}^{4} (x^{2} + 6) d x

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

\int_{- 1}^{1} x^{2} + x + 2 d x

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