What is the integral of 2t^3(t^4-3)^5 ?

The integral of 2t^3(t^4-3)^5 is 1/12 (t^4-3)^6+C

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integral of 2t^3(t^4-3)^5

\int 2 t^{3} (t^{4} - 3)^{5} d t

\frac{1}{12} (t^{4} - 3)^{6} + C

Solution steps

\int 2 t^{3} (t^{4} - 3)^{5} d t

Take the constant out:

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

= 2 \cdot \int t^{3} (t^{4} - 3)^{5} d t

Apply u-substitution

: \int \frac{u ^{5}}{4} d u

= 2 \cdot \int \frac{u ^{5}}{4} d u

Take the constant out:

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

= 2 \cdot \frac{1}{4} \cdot \int u^{5} d u

Apply the Power Rule

: \frac{u ^{6}}{6}

= 2 \cdot \frac{1}{4} \cdot \frac{u ^{6}}{6}

Substitute back

u = t^{4} - 3

= 2 \cdot \frac{1}{4} \cdot \frac{( t ^{4} - 3 ) ^{6}}{6}

Simplify

2 \cdot \frac{1}{4} \cdot \frac{( t ^{4} - 3 ) ^{6}}{6} : \frac{1}{12} (t^{4} - 3)^{6}

= \frac{1}{12} (t^{4} - 3)^{6}

Add a constant to the solution

= \frac{1}{12} (t^{4} - 3)^{6} + C

g (x) = \frac{arcsin ( 3 x )}{x}

f (x) = (x + 2)^{2}

f (x) = \frac{6 x ^{2} + 6 x + 2}{x}

x \to - \infty lim (- 3 x^{3} + x)

f (x) = 2 x^{2}, at x = 1

What is the integral of 2t^3(t^4-3)^5 ?
The integral of 2t^3(t^4-3)^5 is 1/12 (t^4-3)^6+C