What is the integral from 1 to 9 of 1/(16+(x+3)^2) ?

The integral from 1 to 9 of 1/(16+(x+3)^2) is 1/4 (arctan(3)-pi/4)

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integral from 1 to 9 of 1/(16+(x+3)^2)

\int_{1}^{9} \frac{1}{16 + ( x + 3 ) ^{2}} d x

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

Decimal

0.11591 \dots

Solution steps

\int_{1}^{9} \frac{1}{16 + ( x + 3 ) ^{2}} d x

Apply u-substitution

: \int_{4}^{12} \frac{1}{16 + u ^{2}} d u

= \int_{4}^{12} \frac{1}{16 + u ^{2}} d u

Apply Integral Substitution

= \int_{1}^{3} \frac{1}{4 ( 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}{4} \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}{4} [arctan (v)]_{1}^{3}

Compute the boundaries:

arctan (3) - \frac{π}{4}

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

\int_{1}^{2} π (2 - x) d x

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\int_{x}^{1} sin^{2} (x) d x

What is the integral from 1 to 9 of 1/(16+(x+3)^2) ?
The integral from 1 to 9 of 1/(16+(x+3)^2) is 1/4 (arctan(3)-pi/4)