What is the integral from 1 to 7 of 1/(9+(x+2)^2) ?

The integral from 1 to 7 of 1/(9+(x+2)^2) is 1/3 (arctan(3)-pi/4)

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integral from 1 to 7 of 1/(9+(x+2)^2)

\int_{1}^{7} \frac{1}{9 + ( x + 2 ) ^{2}} d x

\frac{1}{3} (arctan (3) - \frac{π}{4})

Decimal Notation

0.15454 \dots

Solution steps

\int_{1}^{7} \frac{1}{9 + ( x + 2 ) ^{2}} d x

Apply u-substitution

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

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

Apply Integral Substitution

= \int_{1}^{3} \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_{1}^{3} \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)]_{1}^{3}

Compute the boundaries:

arctan (3) - \frac{π}{4}

= \frac{1}{3} (arctan (3) - \frac{π}{4})

\int (2 x^{3} - 3 x + 5) d x

\int x cot (x^{2}) d x

y^{^{'}} + y = e^{- t}

n = 1 \sum \infty \frac{n - 5}{n} + 5

\int \frac{x ^{2}}{x ^{2} + 6} d x

What is the integral from 1 to 7 of 1/(9+(x+2)^2) ?
The integral from 1 to 7 of 1/(9+(x+2)^2) is 1/3 (arctan(3)-pi/4)