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

The integral of t/(sqrt(t^2-1)) is sqrt(t^2-1)+C

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

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

t^{2} - 1 + C

Solution steps

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

Apply u-substitution

: \int \frac{1}{2 u} d u

= \int \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 \frac{1}{u} d u

Apply the Power Rule

: 2 u^{\frac{1}{2}}

= \frac{1}{2} \cdot 2 u^{\frac{1}{2}}

Substitute back

u = t^{2} - 1

= \frac{1}{2} \cdot 2 (t^{2} - 1)^{\frac{1}{2}}

Simplify

\frac{1}{2} \cdot 2 (t^{2} - 1)^{\frac{1}{2}} : t^{2} - 1

= t^{2} - 1

Add a constant to the solution

= t^{2} - 1 + C

f (x) = \frac{4}{x}, at 9 \frac{4}{9}

\int \frac{1}{2 x ^{2} + 2 x + 5} d x

\frac{\partial}{\partial x} (x^{\frac{3}{4}} \cdot y^{- 4})

\frac{d}{d x} (4 arcsin (x^{4}))

n = 0 \sum \infty \frac{6 ^{n}}{7 ^{n}}

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