What is the integral of x(x+1)^9 ?

The integral of x(x+1)^9 is 1/110 (x+1)^{10}(10x-1)+C

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integral of x(x+1)^9

\int x (x + 1)^{9} d x

\frac{1}{110} (x + 1)^{10} (10 x - 1) + C

Solution steps

\int x (x + 1)^{9} d x

Apply u-substitution

Take the constant out:

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

Expand

Apply the Sum Rule:

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

\int u^{\frac{2}{9}} d u = \frac{9}{11} u^{\frac{11}{9}}

= \frac{1}{9} (\frac{9}{11} u^{\frac{11}{9}} - \frac{9}{10} u^{\frac{10}{9}})

Substitute back

u = (x + 1)^{9}

= \frac{1}{9} (\frac{9}{11} ((x + 1)^{9})^{\frac{11}{9}} - \frac{9}{10} ((x + 1)^{9})^{\frac{10}{9}})

Simplify

\frac{1}{9} (\frac{9}{11} ((x + 1)^{9})^{\frac{11}{9}} - \frac{9}{10} ((x + 1)^{9})^{\frac{10}{9}}) : \frac{1}{110} (x + 1)^{10} (10 x - 1)

= \frac{1}{110} (x + 1)^{10} (10 x - 1)

Add a constant to the solution

= \frac{1}{110} (x + 1)^{10} (10 x - 1) + C

\int 2 2 x - 1 d x

\int \frac{5}{( 3 x - 1 ) ^{5}} d x

y = tan (arcsin (x))

\frac{\partial}{\partial x} (4 x^{2} y^{3} - 4)

y^{^{'}} = 0.2 ln (\frac{3000}{y}) y, y (0) = 1000

What is the integral of x(x+1)^9 ?
The integral of x(x+1)^9 is 1/110 (x+1)^{10}(10x-1)+C