What is the integral from-2 to 1 of x/(sqrt(2x^2+1)) ?

The integral from-2 to 1 of x/(sqrt(2x^2+1)) is -(3-sqrt(3))/2

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integral from-2 to 1 of x/(sqrt(2x^2+1))

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

- \frac{3 - 3}{2}

Decimal Notation

- 0.63397 \dots

Solution steps

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

Apply u-substitution

: \int_{9}^{3} \frac{1}{4 u} d u

= \int_{9}^{3} \frac{1}{4 u} d u

\int_{a}^{b} f (x) d x = - \int_{b}^{a} f (x) d x, a < b

= - \int_{3}^{9} \frac{1}{4 u} d u

Take the constant out:

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

= - \frac{1}{4} \cdot \int_{3}^{9} \frac{1}{u} d u

Apply the Power Rule

: [2 u^{\frac{1}{2}}]_{3}^{9}

= - \frac{1}{4} [2 u^{\frac{1}{2}}]_{3}^{9}

Apply radical rule:

= - \frac{1}{4} [2 u]_{3}^{9}

Compute the boundaries:

6 - 23

= - \frac{1}{4} (6 - 23)

Simplify

= - \frac{3 - 3}{2}

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What is the integral from-2 to 1 of x/(sqrt(2x^2+1)) ?
The integral from-2 to 1 of x/(sqrt(2x^2+1)) is -(3-sqrt(3))/2