What is the integral of (x^2-7)^4(2x) ?

The integral of (x^2-7)^4(2x) is 1/5 (x^2-7)^5+C

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integral of (x^2-7)^4(2x)

\int (x^{2} - 7)^{4} (2 x) d x

\frac{1}{5} (x^{2} - 7)^{5} + C

Solution steps

\int (x^{2} - 7)^{4} \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} - 7)^{4} x d x

Apply u-substitution

: \int \frac{u ^{4}}{2} d u

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

Apply the Power Rule

: \frac{u ^{5}}{5}

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

Substitute back

u = x^{2} - 7

= 2 \cdot \frac{1}{2} \cdot \frac{( x ^{2} - 7 ) ^{5}}{5}

Simplify

2 \cdot \frac{1}{2} \cdot \frac{( x ^{2} - 7 ) ^{5}}{5} : \frac{1}{5} (x^{2} - 7)^{5}

= \frac{1}{5} (x^{2} - 7)^{5}

Add a constant to the solution

= \frac{1}{5} (x^{2} - 7)^{5} + C

t^{2} y^{^{''}} + 19 t y^{^{'}} + 106 y = 0

x \to 0 lim (tan (\frac{1}{x}))

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

\frac{\partial}{\partial p} (\frac{( p + q )}{p + r})

x \to 0 lim (\frac{sin ( 4 x )}{cos ( 5 x )})

What is the integral of (x^2-7)^4(2x) ?
The integral of (x^2-7)^4(2x) is 1/5 (x^2-7)^5+C