Problem: Definite Integral with (a^2+x^2)^2

[Vect]

\(\displaystyle \displaystyle \int\limits_0^1 {{2 \over {{{(1 + {x^2})}^2}}}dx} \)

Is there any easier way to approach this problem, rather than using the following formula?

\(\displaystyle \displaystyle \int {{1 \over {{{({a^2} + {x^2})}^2}}}} dx = {{{\textstyle{{ax} \over {{a^2} + {x^2}}}} + \arctan ({\textstyle{x \over a}})} \over {2{a^3}}} + C \)

Many thanks in advance.

 

[Deleted member 4993]

\(\displaystyle \displaystyle \int\limits_0^1 {{2 \over {{{(1 + {x^2})}^2}}}dx} \)

Is there any easier way to approach this problem, rather than using the following formula?

\(\displaystyle \displaystyle \int {{1 \over {{{({a^2} + {x^2})}^2}}}} dx = {{{\textstyle{{ax} \over {{a^2} + {x^2}}}} + \arctan ({\textstyle{x \over a}})} \over {2{a^3}}} + C \)

Many thanks in advance.

You can derive the formula by sustituting:

x = tan(Θ)

dx = sec²(Θ) dΘ

\(\displaystyle \displaystyle \int\limits_0^1 {{2 \over {{{(1 + {x^2})}^2}}}dx} \)

\(\displaystyle \displaystyle {= \ \int\limits_0^{\frac{\pi}{4}} {{2 \over {{{sec^4(\theta)}}}}sec^2(\theta) \ d\theta}} \)

and continue.....

 

[Vect]

Solved step-by-step as follows:

\(\displaystyle \displaystyle \eqalign{ & I = \int\limits_0^1 {{2 \over {{{(1 + {x^2})}^2}}}dx} \cr
& x = \tan (y) \cr
& dx = {1 \over {{{\cos }^2}(y)}}dy \cr
& x = 1 \to y = \arctan (1) = {\pi \over 4} \cr
& x = 0 \to y = \arctan (0) = 0 \cr
& I = \int\limits_0^{{\pi \over 4}} {{2 \over {{{({{\tan }^2}(y) + 1)}^2}}} \cdot {1 \over {{{\cos }^2}(y)}}dy} \cr
& \cr
& I = \int\limits_0^{{\pi \over 4}} {{2 \over {{{({1 \over {{{\cos }^2}(y)}})}^2}}} \cdot {1 \over {{{\cos }^2}(y)}}dy} = 2\int\limits_0^{{\pi \over 4}} {{{{{\cos }^4}(y)} \over {{{\cos }^2}(y)}}dy} \cr
& I = \left. {{2 \over 2}\left[ {y
+ \sin (y)\cos (y)} \right]} \right|_0^{{\pi \over 4}} = {\pi \over 4} + \sin ({\pi \over 4})\cos ({\pi \over 4}) - 0 = {\pi \over 4} + {{\sqrt 2 } \over 2} \cdot {{\sqrt 2 } \over 2} = {\pi \over 4} + {1 \over 2} \cr} \)