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

= 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

= 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}

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

x \to 0 lim (- \frac{ln ( 1 + 6 x )}{x})

f (t) = \frac{t sin ( t )}{1 + t}

\int \frac{2 - x}{x} d x

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