What is the integral of (t^3)/(t^4+1) ?

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

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integral of (t^3)/(t^4+1)

\int \frac{t ^{3}}{t ^{4} + 1} d t

\frac{1}{4} ln t^{4} + 1 + C

Solution steps

\int \frac{t ^{3}}{t ^{4} + 1} 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} + 1

= \frac{1}{4} ln t^{4} + 1

Add a constant to the solution

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

\int \frac{2}{cos ( x ) - 1} d x

\int \frac{y - 1}{y} d y

\frac{d}{d x} (lo g_{2} (2 x + 1))

x^{2} + 2, x + 2

6 e^{- 3 x}

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