"Evaluate the definite integral. ∫1719+(x+2)2dx" solved and explained

High School
Calculus and Analysis
Calculus 2
Applications of integrals

Oberlaudacu

Answered question

2021-12-12

Evaluate the definite integral.

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

Answer & Explanation

Stella Calderon

Beginner2021-12-13Added 35 answers

Step 1
Given integral is

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

Step 2
Applying u-substitution,
u=x+2

\Rightarrow d u = d x

limits,
when x=1, u=3
when x=7, u=9
the integral becomes

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

again substituting u=3v,

\Rightarrow d u = 3 d v

limits, when u=3, v=1
when u=9,v=3
now the integral becomes,

\int_{1}^{3} \frac{3}{9 (v^{2} + 1)} d v

= \int_{1}^{3} \frac{1}{3 (v^{2} + 1)} d v

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

= \frac{1}{3} {[\tan^{- 1} v]}_{1}^{3}

= \frac{1}{3} [\tan^{- 1} (3) - \frac{π}{4}]

Step 3
Therefore,

\int_{1}^{7} \frac{1}{9 + {(x + 2)}^{2}} d x = \frac{1}{3} [\tan^{- 1} (3) - \frac{π}{4}]

=0.1545

Shawn Kim

Beginner2021-12-14Added 25 answers

Given:
$\int_{1}^{7} \frac{1}{9 + {(x + 2)}^{2}} d x$
The integration process can be simplified by making a change of variables:
t=x+2
Then the original integral can be written as follows:
$\int \frac{1}{t^{2} + 9} d t$
This is a tabular integral:
$\int \frac{1}{t^{2} + 9} d t = \frac{\arctan (\frac{t}{3})}{3} + C$
To write down the final answer, it remains to substitute x + 2 instead of t.
$\frac{\arctan (\frac{x}{3} + \frac{2}{3})}{3} + C$
Let's calculate a definite integral:
$\int_{1}^{7} \frac{1}{9 + {(x + 2)}^{2}} d x = (\frac{\arctan (\frac{x}{3} + \frac{2}{3})}{3}) ∣_{1}^{7}$
$F (7) = \frac{0}{3}$
$F (1) = \frac{π}{12}$
$I = \frac{0}{3} - (\frac{π}{12}) = \frac{0 \cdot \arctan}{3} 3 - \frac{π}{12}$

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