What is the integral from 1 to 7 of 1/(36+(x+5)^2) ?

The integral from 1 to 7 of 1/(36+(x+5)^2) is 1/6 (arctan(2)-pi/4)

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+5)^2)

\int_{1}^{7} \frac{1}{36 + ( x + 5 ) ^{2}} d x

\frac{1}{6} (arctan (2) - \frac{π}{4})

Decimal

0.05362 \dots

Solution steps

\int_{1}^{7} \frac{1}{36 + ( x + 5 ) ^{2}} d x

Apply u-substitution

: \int_{6}^{12} \frac{1}{36 + u ^{2}} d u

= \int_{6}^{12} \frac{1}{36 + u ^{2}} d u

Apply Integral Substitution

= \int_{1}^{2} \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_{1}^{2} \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)]_{1}^{2}

Compute the boundaries:

arctan (2) - \frac{π}{4}

= \frac{1}{6} (arctan (2) - \frac{π}{4})

\frac{d}{d x} (\frac{x ^{2} - 4}{x ^{2} + 4})

\int (- 2 x^{4} - 4 x^{2} - 5) d x

\int 2 cos (\frac{x}{2}) d x

\frac{\partial}{\partial y} (2 ln (x) + ln (y))

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

What is the integral from 1 to 7 of 1/(36+(x+5)^2) ?
The integral from 1 to 7 of 1/(36+(x+5)^2) is 1/6 (arctan(2)-pi/4)