What is the integral of (x^2+2x+1)^{20} ?

The integral of (x^2+2x+1)^{20} is ((x+1)^{41})/(41)+C

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integral of (x^2+2x+1)^{20}

\int (x^{2} + 2 x + 1)^{20} d x

\frac{( x + 1 ) ^{41}}{41} + C

Solution steps

\int (x^{2} + 2 x + 1)^{20} d x

Complete the square

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

= \int ((x + 1)^{2})^{20} d x

Apply u-substitution

: \int u^{40} d u

= \int u^{40} d u

Apply the Power Rule

: \frac{u ^{41}}{41}

= \frac{u ^{41}}{41}

Substitute back

u = x + 1

= \frac{( x + 1 ) ^{41}}{41}

Add a constant to the solution

= \frac{( x + 1 ) ^{41}}{41} + C

\frac{d y}{d t} = \frac{t}{y}

\int \frac{1}{x + x ^{\frac{1}{3}}} d x

\frac{d}{d x} (arctan (\frac{1 + x}{1 - x}))

f (x) = x^{3}, (- 2, - 8)

What is the integral of (x^2+2x+1)^{20} ?
The integral of (x^2+2x+1)^{20} is ((x+1)^{41})/(41)+C