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Popular >

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

Solution

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

Solution

\frac{2 arctan ( \frac{1}{2} ) + sin ( 2 arctan ( \frac{1}{2} ) )}{8 2}

+1

Solution steps

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

Apply Integral Substitution

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

Take the constant out:

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

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

Factor

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

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

Take the constant out:

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

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

Apply Trigonometric Substitution

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

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

Rewrite using trig identities

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

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

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

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

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

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

Simplify

2 \frac{1}{4} \cdot \frac{1}{2} arctan (\frac{1}{2}) + \frac{sin ( 2 arctan ( \frac{1}{2} ) )}{2} : \frac{2 arctan ( \frac{1}{2} ) + sin ( 2 arctan ( \frac{1}{2} ) )}{8 2}

= \frac{2 arctan ( \frac{1}{2} ) + sin ( 2 arctan ( \frac{1}{2} ) )}{8 2}

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