What is the integral from-1 to 2 of 2/(y^2+2y+10) ?

The integral from-1 to 2 of 2/(y^2+2y+10) is pi/6

English

Español

Português

Français

Deutsch

Italiano

Русский

中文(简体)

한국어

日本語

Tiếng Việt

עברית

العربية

Popular >

integral from-1 to 2 of 2/(y^2+2y+10)

\int_{- 1}^{2} \frac{2}{y ^{2} + 2 y + 10} d y

\frac{π}{6}

Decimal Notation

0.52359 \dots

Solution steps

\int_{- 1}^{2} \frac{2}{y ^{2} + 2 y + 10} d y

Take the constant out:

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

= 2 \cdot \int_{- 1}^{2} \frac{1}{y ^{2} + 2 y + 10} d y

Complete the square

y^{2} + 2 y + 10 : (y + 1)^{2} + 9

= 2 \cdot \int_{- 1}^{2} \frac{1}{( y + 1 ) ^{2} + 9} d y

Apply u-substitution

: \int_{0}^{3} \frac{1}{u ^{2} + 9} d u

= 2 \cdot \int_{0}^{3} \frac{1}{u ^{2} + 9} d u

Apply Integral Substitution

= 2 \cdot \int_{0}^{1} \frac{1}{3 ( v ^{2} + 1 )} d v

Take the constant out:

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

= 2 \cdot \frac{1}{3} \cdot \int_{0}^{1} \frac{1}{v ^{2} + 1} d v

Use the common integral:

\int \frac{1}{v ^{2} + 1} d v = arctan (v)

= 2 \cdot \frac{1}{3} [arctan (v)]_{0}^{1}

Simplify

2 \cdot \frac{1}{3} [arctan (v)]_{0}^{1} : \frac{2}{3} [arctan (v)]_{0}^{1}

= \frac{2}{3} [arctan (v)]_{0}^{1}

Compute the boundaries:

\frac{π}{4}

= \frac{2}{3} \cdot \frac{π}{4}

Simplify

= \frac{π}{6}

\int_{1}^{2} \frac{1}{( 2 x + 3 y ) ^{2}} d y

\int_{1}^{e^{9}} \frac{( ln ( x ) ) ^{2}}{x} d x

\int_{- 2}^{1} (2 x^{2} - 6 x + 5) d x

\int_{0}^{8} ∣ 5 - x ∣ d x

\int_{0}^{x} \frac{2 a ^{2}}{x ^{3}} d x

What is the integral from-1 to 2 of 2/(y^2+2y+10) ?
The integral from-1 to 2 of 2/(y^2+2y+10) is pi/6