What is the integral of (t^3)/(t^2+1) ?

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

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integral of (t^3)/(t^2+1)

integral

\frac{t ^{3}}{t ^{2} + 1}

\frac{1}{2} (t^{2} + 1 - ln t^{2} + 1) + C

Solution steps

\int \frac{t ^{3}}{t ^{2} + 1} d t

Apply u-substitution

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

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

Take the constant out:

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

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

Expand

\frac{u - 1}{u} : 1 - \frac{1}{u}

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

Apply the Sum Rule:

\int f (x) \pm g (x) d x = \int f (x) d x \pm \int g (x) d x

= \frac{1}{2} (\int 1 d u - \int \frac{1}{u} d u)

\int 1 d u = u

\int \frac{1}{u} d u = ln ∣ u ∣

= \frac{1}{2} (u - ln ∣ u ∣)

Substitute back

u = t^{2} + 1

= \frac{1}{2} (t^{2} + 1 - ln t^{2} + 1)

Add a constant to the solution

= \frac{1}{2} (t^{2} + 1 - ln t^{2} + 1) + C

\frac{1}{( x ^{2} - 6 x + 5 )}

\frac{5}{x + 1}

\frac{( 1 + x ) ^{2}}{x ^{2}}

\frac{( 1 - x ^{2} )}{( e ^{2} x )}

\frac{x ^{- 2}}{3}

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