What is the integral from 0 to 1 of x^3sqrt(4x^2+1) ?

The integral from 0 to 1 of x^3sqrt(4x^2+1) is (25sqrt(5)+1)/(120)

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integral from 0 to 1 of x^3sqrt(4x^2+1)

\int_{0}^{1} x^{3} 4 x^{2} + 1 d x

\frac{25 5 + 1}{120}

Decimal

0.47418 \dots

Solution steps

\int_{0}^{1} x^{3} 4 x^{2} + 1 d x

Apply u-substitution

: \int_{1}^{5} \frac{u ( u - 1 )}{32} d u

= \int_{1}^{5} \frac{u ( u - 1 )}{32} d u

Take the constant out:

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

= \frac{1}{32} \cdot \int_{1}^{5} u (u - 1) d u

Expand

u (u - 1) : u^{\frac{3}{2}} - u

= \frac{1}{32} \cdot \int_{1}^{5} u^{\frac{3}{2}} - u 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

= \frac{1}{32} (\int_{1}^{5} u^{\frac{3}{2}} d u - \int_{1}^{5} u d u)

\int_{1}^{5} u^{\frac{3}{2}} d u = 105 - \frac{2}{5}

\int_{1}^{5} u d u = \frac{10 5 - 2}{3}

= \frac{1}{32} (105 - \frac{2}{5} - \frac{10 5 - 2}{3})

Simplify

\frac{1}{32} (105 - \frac{2}{5} - \frac{10 5 - 2}{3}) : \frac{25 5 + 1}{120}

= \frac{25 5 + 1}{120}

\int_{0}^{5 π} sin (x) d x

\int_{x}^{3} sin (t^{4}) d t

\int_{0}^{2} x^{2} - 2 x + 1 d x

\int_{0}^{π} e^{x} cos (e^{x}) d x

\int_{0}^{19 π} sin (x) d x

What is the integral from 0 to 1 of x^3sqrt(4x^2+1) ?
The integral from 0 to 1 of x^3sqrt(4x^2+1) is (25sqrt(5)+1)/(120)