What is the integral of t(t^2+1)^4 ?

The integral of t(t^2+1)^4 is 1/10 (t^2+1)^5+C

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integral of t(t^2+1)^4

\int t (t^{2} + 1)^{4} d t

\frac{1}{10} (t^{2} + 1)^{5} + C

Solution steps

\int t (t^{2} + 1)^{4} d t

Apply u-substitution

: \int \frac{u ^{4}}{2} d u

= \int \frac{u ^{4}}{2} 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 u^{4} d u

Apply the Power Rule

: \frac{u ^{5}}{5}

= \frac{1}{2} \cdot \frac{u ^{5}}{5}

Substitute back

u = t^{2} + 1

= \frac{1}{2} \cdot \frac{( t ^{2} + 1 ) ^{5}}{5}

Simplify

\frac{1}{2} \cdot \frac{( t ^{2} + 1 ) ^{5}}{5} : \frac{1}{10} (t^{2} + 1)^{5}

= \frac{1}{10} (t^{2} + 1)^{5}

Add a constant to the solution

= \frac{1}{10} (t^{2} + 1)^{5} + C

n = 1 \sum \infty cos^{n} (0.6)

\frac{d}{d x} (\frac{1}{x ^{3} + 4 x ^{2}} e^{x c o s (x)})

x e^{- x}

x \to - 5 lim (x - 5)

X^{2} - 2 X, (0, 3)

What is the integral of t(t^2+1)^4 ?
The integral of t(t^2+1)^4 is 1/10 (t^2+1)^5+C