What is the integral from-2 to 1 of x/(sqrt(6x^2+1)) ?

The integral from-2 to 1 of x/(sqrt(6x^2+1)) is -(5-sqrt(7))/6

English

Español

Português

Français

Deutsch

Italiano

Русский

中文(简体)

한국어

日本語

Tiếng Việt

עברית

العربية

Popular >

integral from-2 to 1 of x/(sqrt(6x^2+1))

\int_{- 2}^{1} \frac{x}{6 x ^{2} + 1} d x

- \frac{5 - 7}{6}

Decimal Notation

- 0.39237 \dots

Solution steps

\int_{- 2}^{1} \frac{x}{6 x ^{2} + 1} d x

Apply u-substitution

: \int_{25}^{7} \frac{1}{12 u} d u

= \int_{25}^{7} \frac{1}{12 u} d u

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

= - \int_{7}^{25} \frac{1}{12 u} d u

Take the constant out:

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

= - \frac{1}{12} \cdot \int_{7}^{25} \frac{1}{u} d u

Apply the Power Rule

: [2 u^{\frac{1}{2}}]_{7}^{25}

= - \frac{1}{12} [2 u^{\frac{1}{2}}]_{7}^{25}

Apply radical rule:

= - \frac{1}{12} [2 u]_{7}^{25}

Compute the boundaries:

10 - 27

= - \frac{1}{12} (10 - 27)

Simplify

= - \frac{5 - 7}{6}

\int_{- l n (2)}^{l n (2)} sinh (x) d x

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

\int_{- 1}^{2} (3 x - 3) d x

\int_{0}^{\frac{π}{2}} (x + 1) cos (2 x) d x

\int_{0}^{2 π} 1 + cos (x) d x

What is the integral from-2 to 1 of x/(sqrt(6x^2+1)) ?
The integral from-2 to 1 of x/(sqrt(6x^2+1)) is -(5-sqrt(7))/6