What is the integral of (t^5)/(t^6+1) ?

The integral of (t^5)/(t^6+1) is 1/6 ln|t^6+1|+C

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integral of (t^5)/(t^6+1)

\int \frac{t ^{5}}{t ^{6} + 1} d t

\frac{1}{6} ln t^{6} + 1 + C

Solution steps

\int \frac{t ^{5}}{t ^{6} + 1} d t

Apply u-substitution

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

= \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

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

Use the common integral:

\int \frac{1}{u} d u = ln (∣ u ∣)

= \frac{1}{6} ln ∣ u ∣

Substitute back

u = t^{6} + 1

= \frac{1}{6} ln t^{6} + 1

Add a constant to the solution

= \frac{1}{6} ln t^{6} + 1 + C

\frac{d}{d x} (tan (\frac{x}{y}))

n = 1 \sum \infty \frac{2}{4 ^{2 n}}

h \to 0 lim (\frac{( \frac{7}{x + h} - \frac{7}{x} )}{h})

x \to 1 lim (ln ∣ x - 1 ∣)

\frac{\partial ^{2}}{\partial x ^{2}} (2 cos (x y^{2}))

What is the integral of (t^5)/(t^6+1) ?
The integral of (t^5)/(t^6+1) is 1/6 ln|t^6+1|+C