What is the integral from 0 to 1 of 1/(x^2+6x+9) ?

The integral from 0 to 1 of 1/(x^2+6x+9) is 1/12

English

Español

Português

Français

Deutsch

Italiano

Русский

中文(简体)

한국어

日本語

Tiếng Việt

עברית

العربية

Popular >

integral from 0 to 1 of 1/(x^2+6x+9)

\int_{0}^{1} \frac{1}{x ^{2} + 6 x + 9} d x

\frac{1}{12}

Decimal

0.08333 \dots

Solution steps

\int_{0}^{1} \frac{1}{x ^{2} + 6 x + 9} d x

Take the partial fraction of

\frac{1}{x ^{2} + 6 x + 9} : \frac{1}{( x + 3 ) ^{2}}

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

Apply u-substitution

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

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

Apply the Power Rule

: [- \frac{1}{u}]_{3}^{4}

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

Compute the boundaries:

\frac{1}{12}

= \frac{1}{12}

\int_{4}^{6} x^{2} + 2 x - 4 d x

\int_{- 1}^{5} x + 3 d x

\int_{0}^{\infty} e^{- 2 x} x d x

\int_{- 2}^{2} \frac{1}{x + 3} d x

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

What is the integral from 0 to 1 of 1/(x^2+6x+9) ?
The integral from 0 to 1 of 1/(x^2+6x+9) is 1/12