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

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

English

Español

Português

Français

Deutsch

Italiano

Русский

中文(简体)

한국어

日本語

Tiếng Việt

עברית

العربية

Popular >

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

\int_{6}^{9} \frac{1}{9 + ( x - 6 ) ^{2}} d x

\frac{π}{12}

Decimal Notation

0.26179 \dots

Solution steps

\int_{6}^{9} \frac{1}{9 + ( x - 6 ) ^{2}} d x

Apply u-substitution

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

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

Apply Integral Substitution

= \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

= \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)

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

Compute the boundaries:

\frac{π}{4}

= \frac{1}{3} \cdot \frac{π}{4}

Simplify

= \frac{π}{12}

\frac{d}{d x} (x + c)

\frac{d}{d x} (x^{4} sin (x) cos (x))

\int_{0}^{1} 6^{x} d x

y = x^{3} - 3 x + 1, (3, 19)

\frac{d}{d x} (I n (x x^{2} - 1))

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