What is the integral from 0 to 1 of (x^2+1)^{10}(2x) ?

The integral from 0 to 1 of (x^2+1)^{10}(2x) is 2047/11

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integral from 0 to 1 of (x^2+1)^{10}(2x)

\int_{0}^{1} (x^{2} + 1)^{10} (2 x) d x

\frac{2047}{11}

Decimal

186.09090 \dots

Solution steps

\int_{0}^{1} (x^{2} + 1)^{10} \cdot 2 x d x

Take the constant out:

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

= 2 \cdot \int_{0}^{1} (x^{2} + 1)^{10} x d x

Apply u-substitution

: \int_{1}^{2} \frac{u ^{10}}{2} d u

= 2 \cdot \int_{1}^{2} \frac{u ^{10}}{2} d u

Take the constant out:

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

= 2 \cdot \frac{1}{2} \cdot \int_{1}^{2} u^{10} d u

Apply the Power Rule

: [\frac{u ^{11}}{11}]_{1}^{2}

= 2 \cdot \frac{1}{2} [\frac{u ^{11}}{11}]_{1}^{2}

Simplify

2 \cdot \frac{1}{2} [\frac{u ^{11}}{11}]_{1}^{2} : [\frac{u ^{11}}{11}]_{1}^{2}

= [\frac{u ^{11}}{11}]_{1}^{2}

Compute the boundaries:

\frac{2047}{11}

= \frac{2047}{11}

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

\int e^{- 10 x^{2}} d x

\int_{- 1}^{2} x^{2} d x

3 tan (θ)

y^{^{'}} + y = x y^{2}, y (0) = \frac{1}{2}

What is the integral from 0 to 1 of (x^2+1)^{10}(2x) ?
The integral from 0 to 1 of (x^2+1)^{10}(2x) is 2047/11