What is the integral from 1 to 7 of 1/(36+(x-1)^2) ?

The integral from 1 to 7 of 1/(36+(x-1)^2) is pi/(24)

English

Español

Português

Français

Deutsch

Italiano

Русский

中文(简体)

한국어

日本語

Tiếng Việt

עברית

العربية

Popular >

integral from 1 to 7 of 1/(36+(x-1)^2)

\int_{1}^{7} \frac{1}{36 + ( x - 1 ) ^{2}} d x

\frac{π}{24}

Decimal Notation

0.13089 \dots

Solution steps

\int_{1}^{7} \frac{1}{36 + ( x - 1 ) ^{2}} d x

Apply u-substitution

: \int_{0}^{6} \frac{1}{36 + u ^{2}} d u

= \int_{0}^{6} \frac{1}{36 + u ^{2}} d u

Apply Integral Substitution

= \int_{0}^{1} \frac{1}{6 ( 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}{6} \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}{6} [arctan (v)]_{0}^{1}

Compute the boundaries:

\frac{π}{4}

= \frac{1}{6} \cdot \frac{π}{4}

Simplify

= \frac{π}{24}

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

sin (7 t)

\frac{\partial}{\partial x} (10 x)

\int_{- 1}^{2} (3 x^{2} - 4 x - 1) d x

\frac{d}{d t} (2 t - t^{2})

What is the integral from 1 to 7 of 1/(36+(x-1)^2) ?
The integral from 1 to 7 of 1/(36+(x-1)^2) is pi/(24)