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

The integral from 1 to 3 of 1/((x^2+3)^2) is pi/(36sqrt(3))

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

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

\frac{π}{36 3}

Decimal

0.05038 \dots

Solution steps

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

Apply Integral Substitution

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

Take the constant out:

\int a \cdot f (x) d x = a \cdot \int f (x) d x

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

Factor

\frac{1}{( 3 u ^{2} + 3 ) ^{2}}

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

Take the constant out:

\int a \cdot f (x) d x = a \cdot \int f (x) d x

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

Apply Trigonometric Substitution

: \int_{\frac{π}{6}}^{\frac{π}{3}} \frac{1}{sec ^{4} ( v ) cos ^{2} ( v )} d v

= 3 \frac{1}{9} \cdot \int_{\frac{π}{6}}^{\frac{π}{3}} \frac{1}{sec ^{4} ( v ) cos ^{2} ( v )} d v

Rewrite using trig identities

= 3 \frac{1}{9} \cdot \int_{\frac{π}{6}}^{\frac{π}{3}} \frac{1 + cos ( 2 v )}{2} d v

Take the constant out:

\int a \cdot f (x) d x = a \cdot \int f (x) d x

= 3 \frac{1}{9} \cdot \frac{1}{2} \cdot \int_{\frac{π}{6}}^{\frac{π}{3}} 1 + cos (2 v) d v

Apply the Sum Rule:

\int f (x) \pm g (x) d x = \int f (x) d x \pm \int g (x) d x

= 3 \frac{1}{9} \cdot \frac{1}{2} (\int_{\frac{π}{6}}^{\frac{π}{3}} 1 d v + \int_{\frac{π}{6}}^{\frac{π}{3}} cos (2 v) d v)

\int_{\frac{π}{6}}^{\frac{π}{3}} 1 d v = \frac{π}{6}

\int_{\frac{π}{6}}^{\frac{π}{3}} cos (2 v) d v = 0

= 3 \frac{1}{9} \cdot \frac{1}{2} (\frac{π}{6} + 0)

Simplify

3 \frac{1}{9} \cdot \frac{1}{2} (\frac{π}{6} + 0) : \frac{π}{36 3}

= \frac{π}{36 3}

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What is the integral from 1 to 3 of 1/((x^2+3)^2) ?
The integral from 1 to 3 of 1/((x^2+3)^2) is pi/(36sqrt(3))