What is the integral of 30x*(x+1)^4 ?

The integral of 30x*(x+1)^4 is 30((x^6)/6+(4x^5)/5+(3x^4)/2+(4x^3)/3+(x^2)/2)+C

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integral of 30x*(x+1)^4

\int 30 x \cdot (x + 1)^{4} d x

30 (\frac{x ^{6}}{6} + \frac{4 x ^{5}}{5} + \frac{3 x ^{4}}{2} + \frac{4 x ^{3}}{3} + \frac{x ^{2}}{2}) + C

Solution steps

\int 30 x (x + 1)^{4} d x

Take the constant out:

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

= 30 \cdot \int x (x + 1)^{4} d x

Expand

x (x + 1)^{4} : x^{5} + 4 x^{4} + 6 x^{3} + 4 x^{2} + x

= 30 \cdot \int x^{5} + 4 x^{4} + 6 x^{3} + 4 x^{2} + x d x

Apply the Sum Rule:

\int f (x) \pm g (x) d x = \int f (x) d x \pm \int g (x) d x

= 30 (\int x^{5} d x + \int 4 x^{4} d x + \int 6 x^{3} d x + \int 4 x^{2} d x + \int x d x)

\int x^{5} d x = \frac{x ^{6}}{6}

\int 4 x^{4} d x = \frac{4 x ^{5}}{5}

\int 6 x^{3} d x = \frac{3 x ^{4}}{2}

\int 4 x^{2} d x = \frac{4 x ^{3}}{3}

\int x d x = \frac{x ^{2}}{2}

= 30 (\frac{x ^{6}}{6} + \frac{4 x ^{5}}{5} + \frac{3 x ^{4}}{2} + \frac{4 x ^{3}}{3} + \frac{x ^{2}}{2})

Add a constant to the solution

= 30 (\frac{x ^{6}}{6} + \frac{4 x ^{5}}{5} + \frac{3 x ^{4}}{2} + \frac{4 x ^{3}}{3} + \frac{x ^{2}}{2}) + C

\frac{d}{d x} (\frac{2 x}{( 1 + x ^{2} ) ^{2}})

\int \frac{x ^{4}}{x ^{4} - 16} d x

y^{^{''}} + 9 y = 2 cos (3 x) + 3 sin (3 x)

f (x) = 4 sin (x), at x = \frac{π}{2}

y y^{^{'}} = y^{2} + 1

What is the integral of 30x*(x+1)^4 ?
The integral of 30x*(x+1)^4 is 30((x^6)/6+(4x^5)/5+(3x^4)/2+(4x^3)/3+(x^2)/2)+C