What is the integral of (t^3+2)/(t^4+8t-3) ?

The integral of (t^3+2)/(t^4+8t-3) is 1/4 ln|t^4+8t-3|+C

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integral of (t^3+2)/(t^4+8t-3)

\int \frac{t ^{3} + 2}{t ^{4} + 8 t - 3} d t

\frac{1}{4} ln t^{4} + 8 t - 3 + C

Solution steps

\int \frac{t ^{3} + 2}{t ^{4} + 8 t - 3} d t

Apply u-substitution

: \int \frac{1}{4 u} d u

= \int \frac{1}{4 u} d u

Take the constant out:

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

= \frac{1}{4} \cdot \int \frac{1}{u} d u

Use the common integral:

\int \frac{1}{u} d u = ln (∣ u ∣)

= \frac{1}{4} ln ∣ u ∣

Substitute back

u = t^{4} + 8 t - 3

= \frac{1}{4} ln t^{4} + 8 t - 3

Add a constant to the solution

= \frac{1}{4} ln t^{4} + 8 t - 3 + C

x \to \infty - lim (e^{x})

\int_{0}^{x} e^{x + y} d y

\frac{x + 1}{x + 2}

\frac{d}{d x} (- 64 sin (2 x))

y = 6 e^{x}, at x = 1

What is the integral of (t^3+2)/(t^4+8t-3) ?
The integral of (t^3+2)/(t^4+8t-3) is 1/4 ln|t^4+8t-3|+C