What is the integral of (1+x)(2x+x^2)^5 ?

The integral of (1+x)(2x+x^2)^5 is 1/12 (2x+x^2)^6+C

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integral of (1+x)(2x+x^2)^5

\int (1 + x) (2 x + x^{2})^{5} d x

\frac{1}{12} (2 x + x^{2})^{6} + C

Solution steps

\int (1 + x) (2 x + x^{2})^{5} d x

Apply u-substitution

: \int \frac{u ^{5}}{2} d u

= \int \frac{u ^{5}}{2} d u

Take the constant out:

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

= \frac{1}{2} \cdot \int u^{5} d u

Apply the Power Rule

: \frac{u ^{6}}{6}

= \frac{1}{2} \cdot \frac{u ^{6}}{6}

Substitute back

u = 2 x + x^{2}

= \frac{1}{2} \cdot \frac{( 2 x + x ^{2} ) ^{6}}{6}

Simplify

\frac{1}{2} \cdot \frac{( 2 x + x ^{2} ) ^{6}}{6} : \frac{1}{12} (2 x + x^{2})^{6}

= \frac{1}{12} (2 x + x^{2})^{6}

Add a constant to the solution

= \frac{1}{12} (2 x + x^{2})^{6} + C

\frac{d}{d x} ((x - 4) (2 x + x^{2}))

f (x) = 3 t^{3} - 18 t^{2} + 2

\frac{d}{d x} (2 x^{\frac{5}{3}} - 5 x^{\frac{4}{3}})

x \to - \infty lim (- x)

\int 8 x (4 x^{2} + 4)^{4} d x

What is the integral of (1+x)(2x+x^2)^5 ?
The integral of (1+x)(2x+x^2)^5 is 1/12 (2x+x^2)^6+C