What is the integral from 0 to 1 of 1/((16+x^2)^2) ?

The integral from 0 to 1 of 1/((16+x^2)^2) is 1/128 arctan(1/4)+1/544

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integral from 0 to 1 of 1/((16+x^2)^2)

\int_{0}^{1} \frac{1}{( 16 + x ^{2} ) ^{2}} d x

\frac{1}{128} arctan (\frac{1}{4}) + \frac{1}{544}

Decimal Notation

0.00375 \dots

Solution steps

\int_{0}^{1} \frac{1}{( 16 + x ^{2} ) ^{2}} d x

Apply Integral Substitution

= \int_{0}^{\frac{1}{4}} \frac{1}{64 ( u ^{2} + 1 ) ^{2}} d u

Take the constant out:

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

= \frac{1}{64} \cdot \int_{0}^{\frac{1}{4}} \frac{1}{( u ^{2} + 1 ) ^{2}} d u

Apply Trigonometric Substitution

: \int_{0}^{a r c t a n (\frac{1}{4})} \frac{1}{sec ^{4} ( v ) cos ^{2} ( v )} d v

= \frac{1}{64} \cdot \int_{0}^{a r c t a n (\frac{1}{4})} \frac{1}{sec ^{4} ( v ) cos ^{2} ( v )} d v

Rewrite using trig identities

= \frac{1}{64} \cdot \int_{0}^{a r c t a n (\frac{1}{4})} \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

= \frac{1}{64} \cdot \frac{1}{2} \cdot \int_{0}^{a r c t a n (\frac{1}{4})} 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

= \frac{1}{64} \cdot \frac{1}{2} (\int_{0}^{a r c t a n (\frac{1}{4})} 1 d v + \int_{0}^{a r c t a n (\frac{1}{4})} cos (2 v) d v)

\int_{0}^{a r c t a n (\frac{1}{4})} 1 d v = arctan (\frac{1}{4})

\int_{0}^{a r c t a n (\frac{1}{4})} cos (2 v) d v = \frac{4}{17}

= \frac{1}{64} \cdot \frac{1}{2} (arctan (\frac{1}{4}) + \frac{4}{17})

Simplify

\frac{1}{64} \cdot \frac{1}{2} (arctan (\frac{1}{4}) + \frac{4}{17}) : \frac{1}{128} arctan (\frac{1}{4}) + \frac{1}{544}

= \frac{1}{128} arctan (\frac{1}{4}) + \frac{1}{544}

x \to 0 lim (\frac{1}{5 + 5})

\int \frac{1}{1 - ( \frac{x}{L} )} d x

\frac{d y}{d x} + 2 y = 1

\frac{d y}{d x} + 3 x^{2} \cdot y = 6 x^{2}

\frac{\partial}{\partial y} (arctan (x y))

What is the integral from 0 to 1 of 1/((16+x^2)^2) ?
The integral from 0 to 1 of 1/((16+x^2)^2) is 1/128 arctan(1/4)+1/544