What is the integral of x(10x+10)^{100} ?

The integral of x(10x+10)^{100} is 10^{100}(((x+1)^{102})/(102)-((x+1)^{101})/(101))+C

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integral of x(10x+10)^{100}

\int x (10 x + 10)^{100} d x

1 0^{100} (\frac{( x + 1 ) ^{102}}{102} - \frac{( x + 1 ) ^{101}}{101}) + C

Solution steps

\int x (10 x + 10)^{100} d x

Factor

x (10 x + 10)^{100} : 1 0^{100} x (x + 1)^{100}

= \int 1 0^{100} x (x + 1)^{100} d x

Take the constant out:

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

= 1 0^{100} \cdot \int x (x + 1)^{100} d x

Apply u-substitution

= 1 0^{100} \cdot \int (u - 1) u^{100} d u

Expand

(u - 1) u^{100} : u^{101} - u^{100}

= 1 0^{100} \cdot \int u^{101} - u^{100} d u

Apply the Sum Rule:

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

= 1 0^{100} (\int u^{101} d u - \int u^{100} d u)

\int u^{101} d u = \frac{u ^{102}}{102}

\int u^{100} d u = \frac{u ^{101}}{101}

= 1 0^{100} (\frac{u ^{102}}{102} - \frac{u ^{101}}{101})

Substitute back

u = x + 1

= 1 0^{100} (\frac{( x + 1 ) ^{102}}{102} - \frac{( x + 1 ) ^{101}}{101})

Add a constant to the solution

= 1 0^{100} (\frac{( x + 1 ) ^{102}}{102} - \frac{( x + 1 ) ^{101}}{101}) + C

What is the integral of x(10x+10)^{100} ?
The integral of x(10x+10)^{100} is 10^{100}(((x+1)^{102})/(102)-((x+1)^{101})/(101))+C