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

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

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

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

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

Solution steps

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

Take the constant out:

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

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

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 t^{2} - 5

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

Simplify

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

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

Add a constant to the solution

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

\frac{\partial}{\partial y} (\frac{2 x}{x ^{2} + y ^{2}})

\int \frac{x ^{2}}{( 16 + x ^{2} ) ^{2}} d x

\int (8 x - 7) \cdot ln (x) d x

x 8 - 2 x^{2}, 0, 2

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