What is the integral of (5x+5)(x-4)(x-4) ?

The integral of (5x+5)(x-4)(x-4) is 5((x^4)/3-(x^4}{12}-4x^3+\frac{x^3)/3+(4x^3)/3+4x^2+16x)+C

English

Español

Português

Français

Deutsch

Italiano

Русский

中文(简体)

한국어

日本語

Tiếng Việt

עברית

العربية

Popular >

integral of (5x+5)(x-4)(x-4)

\int (5 x + 5) (x - 4) (x - 4) d x

5 (\frac{x ^{4}}{3} - \frac{x ^{4}}{12} - 4 x^{3} + \frac{x ^{3}}{3} + \frac{4 x ^{3}}{3} + 4 x^{2} + 16 x) + C

Solution steps

\int (5 x + 5) (x - 4) (x - 4) d x

(5 x + 5) (x - 4) (x - 4) = 5 (x + 1) (x - 4) (x - 4)

= \int 5 (x + 1) (x - 4) (x - 4) d x

Take the constant out:

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

= 5 \cdot \int (x + 1) (x - 4) (x - 4) d x

Simplify

(x + 1) (x - 4) (x - 4) : (x + 1) (x - 4)^{2}

= 5 \cdot \int (x + 1) (x - 4)^{2} d x

Apply Integration By Parts

: (x + 1) (\frac{x ^{3}}{3} - 4 x^{2} + 16 x) - \int \frac{x ^{3}}{3} - 4 x^{2} + 16 x d x

= 5 ((x + 1) (\frac{x ^{3}}{3} - 4 x^{2} + 16 x) - \int \frac{x ^{3}}{3} - 4 x^{2} + 16 x d x)

\int \frac{x ^{3}}{3} - 4 x^{2} + 16 x d x = \frac{x ^{4}}{12} - \frac{4 x ^{3}}{3} + 8 x^{2}

= 5 ((x + 1) (\frac{x ^{3}}{3} - 4 x^{2} + 16 x) - (\frac{x ^{4}}{12} - \frac{4 x ^{3}}{3} + 8 x^{2}))

Expand

(x + 1) (\frac{x ^{3}}{3} - 4 x^{2} + 16 x) - (\frac{x ^{4}}{12} - \frac{4 x ^{3}}{3} + 8 x^{2}) : \frac{x ^{4}}{3} - \frac{x ^{4}}{12} - 4 x^{3} + \frac{x ^{3}}{3} + \frac{4 x ^{3}}{3} + 4 x^{2} + 16 x

= 5 (\frac{x ^{4}}{3} - \frac{x ^{4}}{12} - 4 x^{3} + \frac{x ^{3}}{3} + \frac{4 x ^{3}}{3} + 4 x^{2} + 16 x)

Add a constant to the solution

= 5 (\frac{x ^{4}}{3} - \frac{x ^{4}}{12} - 4 x^{3} + \frac{x ^{3}}{3} + \frac{4 x ^{3}}{3} + 4 x^{2} + 16 x) + C

\int 9 t^{2} + \frac{1}{t ^{3}} d t

y^{^{''}} + y^{^{'}} - 2 y = 6 e^{x}

4 y^{^{''}} + 8 y^{^{'}} + 3 y = 0

\frac{5 s}{( s ^{2} + 4 ) \cdot ( s + 7 )}

f (x) = (\frac{5 x ^{2} + 6 x + 1}{3 x - 5})^{\frac{3}{2}}

What is the integral of (5x+5)(x-4)(x-4) ?
The integral of (5x+5)(x-4)(x-4) is 5((x^4)/3-(x^4}{12}-4x^3+\frac{x^3)/3+(4x^3)/3+4x^2+16x)+C