What is the integral from 1 to 6 of 1/(25+(x+4)^2) ?

The integral from 1 to 6 of 1/(25+(x+4)^2) is 1/5 (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 6 of 1/(25+(x+4)^2)

\int_{1}^{6} \frac{1}{25 + ( x + 4 ) ^{2}} d x

\frac{1}{5} (arctan (2) - \frac{π}{4})

Decimal

0.06435 \dots

Solution steps

\int_{1}^{6} \frac{1}{25 + ( x + 4 ) ^{2}} d x

Apply u-substitution

: \int_{5}^{10} \frac{1}{25 + u ^{2}} d u

= \int_{5}^{10} \frac{1}{25 + u ^{2}} d u

Apply Integral Substitution

= \int_{1}^{2} \frac{1}{5 ( 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}{5} \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}{5} [arctan (v)]_{1}^{2}

Compute the boundaries:

arctan (2) - \frac{π}{4}

= \frac{1}{5} (arctan (2) - \frac{π}{4})

\int_{5}^{9} \frac{1}{x - 5} d x

\int_{0}^{4} (x^{2} - 4) d x

\int_{2}^{\infty} \frac{1}{ln ( n )}

\int_{0}^{\frac{π}{4}} \frac{1}{x ^{2} + 1} d x

\int_{7}^{9} x^{2} + 2 x - 6 d x

What is the integral from 1 to 6 of 1/(25+(x+4)^2) ?
The integral from 1 to 6 of 1/(25+(x+4)^2) is 1/5 (arctan(2)-pi/4)