What is the integral of t/(a+bt^2) ?

The integral of t/(a+bt^2) is 1/(2b)ln|a+bt^2|+C

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integral of t/(a+bt^2)

\int \frac{t}{a + b t ^{2}} d t

\frac{1}{2 b} ln a + b t^{2} + C

Solution steps

\int \frac{t}{a + b t ^{2}} d t

Apply u-substitution

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

= \int \frac{1}{2 b u} d u

Take the constant out:

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

= \frac{1}{2 b} \cdot \int \frac{1}{u} d u

Use the common integral:

\int \frac{1}{u} d u = ln (∣ u ∣)

= \frac{1}{2 b} ln ∣ u ∣

Substitute back

u = a + b t^{2}

= \frac{1}{2 b} ln a + b t^{2}

Add a constant to the solution

= \frac{1}{2 b} ln a + b t^{2} + C

\int 9 e^{- 9 x} d x

sin^{2} (x), sin^{3} (x)

\frac{d}{d x} (sin (\frac{1}{x}) x^{3})

g (x) = ln (x^{2}), (e, 2)

\int_{0}^{8} t + 2 t^{3} d t

What is the integral of t/(a+bt^2) ?
The integral of t/(a+bt^2) is 1/(2b)ln|a+bt^2|+C