What is the integral of (6t)/(8-3t^2) ?

The integral of (6t)/(8-3t^2) is -ln|8-3t^2|+C

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integral of (6t)/(8-3t^2)

\int \frac{6 t}{8 - 3 t ^{2}} d t

- ln 8 - 3 t^{2} + C

Solution steps

\int \frac{6 t}{8 - 3 t ^{2}} d t

Take the constant out:

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

= 6 \cdot \int \frac{t}{8 - 3 t ^{2}} d t

Apply u-substitution

: \int - \frac{1}{6 u} d u

= 6 \cdot \int - \frac{1}{6 u} d u

Take the constant out:

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

= 6 (- \frac{1}{6} \cdot \int \frac{1}{u} d u)

Use the common integral:

\int \frac{1}{u} d u = ln (∣ u ∣)

= 6 (- \frac{1}{6} ln ∣ u ∣)

Substitute back

u = 8 - 3 t^{2}

= 6 (- \frac{1}{6} ln 8 - 3 t^{2})

Simplify

6 (- \frac{1}{6} ln 8 - 3 t^{2}) : - ln 8 - 3 t^{2}

= - ln 8 - 3 t^{2}

Add a constant to the solution

= - ln 8 - 3 t^{2} + C

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What is the integral of (6t)/(8-3t^2) ?
The integral of (6t)/(8-3t^2) is -ln|8-3t^2|+C