What is the integral of 28x(7x^2-5)^3 ?

The integral of 28x(7x^2-5)^3 is 1/2 (7x^2-5)^4+C

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integral of 28x(7x^2-5)^3

\int 28 x (7 x^{2} - 5)^{3} d x

\frac{1}{2} (7 x^{2} - 5)^{4} + C

Solution steps

\int 28 x (7 x^{2} - 5)^{3} d x

Take the constant out:

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

= 28 \cdot \int x (7 x^{2} - 5)^{3} d x

Apply u-substitution

: \int \frac{u ^{3}}{14} d u

= 28 \cdot \int \frac{u ^{3}}{14} d u

Take the constant out:

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

= 28 \cdot \frac{1}{14} \cdot \int u^{3} d u

Apply the Power Rule

: \frac{u ^{4}}{4}

= 28 \cdot \frac{1}{14} \cdot \frac{u ^{4}}{4}

Substitute back

u = 7 x^{2} - 5

= 28 \cdot \frac{1}{14} \cdot \frac{( 7 x ^{2} - 5 ) ^{4}}{4}

Simplify

28 \cdot \frac{1}{14} \cdot \frac{( 7 x ^{2} - 5 ) ^{4}}{4} : \frac{1}{2} (7 x^{2} - 5)^{4}

= \frac{1}{2} (7 x^{2} - 5)^{4}

Add a constant to the solution

= \frac{1}{2} (7 x^{2} - 5)^{4} + C

\int a e^{- a t} d t

3 x^{2} e^{x} + e^{x} (x^{3} + 4)

\int x^{5} e^{x^{6}} d x

(- 3.5) (2.1)

\int_{0}^{2} \frac{x}{2} d x

What is the integral of 28x(7x^2-5)^3 ?
The integral of 28x(7x^2-5)^3 is 1/2 (7x^2-5)^4+C