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

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

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

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

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

Solution steps

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

Take the constant out:

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

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

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 = x^{4} - 3

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

Simplify

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

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

Add a constant to the solution

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

\frac{x ^{2}}{x + 9}

\frac{d}{d x} (\frac{x ^{3} - 4 x ^{3}}{2 x})

n = 1 \sum \infty 9^{n} x^{n} n!

x \to 100 lim (7)

\frac{\partial}{\partial x} (e^{a x + b y})

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