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

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

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

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

\frac{1}{12} (x^{2} - 3)^{6} + C

Solution steps

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

Apply u-substitution

: \int \frac{u ^{5}}{2} d u

= \int \frac{u ^{5}}{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^{5} d u

Apply the Power Rule

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

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

Substitute back

u = x^{2} - 3

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

Simplify

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

= \frac{1}{12} (x^{2} - 3)^{6}

Add a constant to the solution

= \frac{1}{12} (x^{2} - 3)^{6} + C

x \to 3 lim (\frac{2}{x + 3})

\int \frac{1}{( 169 + x ^{2} ) ^{\frac{3}{2}}} d x

y = x^{6} + 3

\frac{d}{d x} (sin (x^{2}) + sin^{2} (x))

\frac{\partial ^{2}}{\partial z ^{2}} (x^{2} y^{4} e^{z} + 3 y^{3} e^{x z} + 2 e^{x yz} + 3 y)

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