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

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

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

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

\frac{1}{6} (x^{2} - 5)^{6} + C

Solution steps

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

Take the constant out:

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

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

Apply u-substitution

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

= 2 \cdot \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

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

Apply the Power Rule

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

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

Substitute back

u = x^{2} - 5

= 2 \cdot \frac{1}{2} \cdot \frac{( x ^{2} - 5 ) ^{6}}{6}

Simplify

2 \cdot \frac{1}{2} \cdot \frac{( x ^{2} - 5 ) ^{6}}{6} : \frac{1}{6} (x^{2} - 5)^{6}

= \frac{1}{6} (x^{2} - 5)^{6}

Add a constant to the solution

= \frac{1}{6} (x^{2} - 5)^{6} + C

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

\frac{x ^{2}}{1 + x}

\int e^{\frac{- t ^{2}}{2}} d t

x \to + (- 2) - lim (\frac{1}{x ^{2} - 4})

\frac{d}{d x} (cos (wt))

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