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

The integral from 0 to 1 of 3x(x^2+1)^4 is 93/10

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

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

\frac{93}{10}

Decimal Notation

9.3

Solution steps

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

Take the constant out:

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

= 3 \cdot \int_{0}^{1} x (x^{2} + 1)^{4} d x

Apply u-substitution

: \int_{1}^{2} \frac{u ^{4}}{2} d u

= 3 \cdot \int_{1}^{2} \frac{u ^{4}}{2} d u

Take the constant out:

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

= 3 \cdot \frac{1}{2} \cdot \int_{1}^{2} u^{4} d u

Apply the Power Rule

: [\frac{u ^{5}}{5}]_{1}^{2}

= 3 \cdot \frac{1}{2} [\frac{u ^{5}}{5}]_{1}^{2}

Simplify

3 \cdot \frac{1}{2} [\frac{u ^{5}}{5}]_{1}^{2} : \frac{3}{2} [\frac{u ^{5}}{5}]_{1}^{2}

= \frac{3}{2} [\frac{u ^{5}}{5}]_{1}^{2}

Compute the boundaries:

\frac{31}{5}

= \frac{3}{2} \cdot \frac{31}{5}

Simplify

= \frac{93}{10}

x^{2} \frac{d y}{d x} + 3 x y = ln (x)

x y = cot (x y)

\int \frac{x}{( z ^{2} + x ^{2} ) ^{\frac{3}{2}}} d x

y = \frac{3}{x}

\int (6) d x

What is the integral from 0 to 1 of 3x(x^2+1)^4 ?
The integral from 0 to 1 of 3x(x^2+1)^4 is 93/10