How would you compute for the definite integral of∫0∞dx(1+x2)4

The method given below works in general for any integral of the form ∫0∞dx(1+x2)nPlug in x=tan⁡(θ). The integral becomes∫0∞dx(1+x2)4=∫0π2sec2(θ)dθ(1+tan2(θ))4=∫0π2cos6(θ)dθ=563412π2=532π

F(u)≡∫0∞dxμ+x2=μ−12∫0∞dx1+x2=12πμ−12F⁗(μ)=−3!∫0∞dx(μ+x2)4}=12πd3μ−12dμ3=12π(−12)(−32)(−52)μ−72Set μ=1 in both members:∫0∞dx(1+x2)4=15π66=532π

The method given below works in general for any integral of the form ∫0∞dx(1+x2)nPlug in x=tan⁡(θ). The integral becomes∫0∞dx(1+x2)4=∫0π2sec2(θ)dθ(1+tan2(θ))4=∫0π2cos6(θ)dθ=563412π2=532π

F(u)≡∫0∞dxμ+x2=μ−12∫0∞dx1+x2=12πμ−12F⁗(μ)=−3!∫0∞dx(μ+x2)4}=12πd3μ−12dμ3=12π(−12)(−32)(−52)μ−72Set μ=1 in both members:∫0∞dx(1+x2)4=15π66=532π

Find out the answer to "How would you compute for the definite integral..."

Zoie Phillips

Answered question

2022-03-30

How would you compute for the definite integral of

\int_{0}^{\infty} \frac{d x}{{(1 + x^{2})}^{4}}

Answer & Explanation

Tristatex9tw

Beginner2022-03-31Added 18 answers

The method given below works in general for any integral of the form

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

Plug in

x = \tan (θ)

. The integral becomes

\int_{0}^{\infty} \frac{d x}{{(1 + x^{2})}^{4}} = \int_{0}^{\frac{π}{2}} \frac{\sec^{2} (θ) d θ}{{(1 + \tan^{2} (θ))}^{4}}

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

= \frac{5}{6} \frac{3}{4} \frac{1}{2} \frac{π}{2}

= \frac{5}{32} π

kaosimqu5t

Beginner2022-04-01Added 10 answers

F (u) \equiv \int_{0}^{\infty} \frac{d x}{μ + x^{2}} = μ^{- \frac{1}{2}} \int_{0}^{\infty} \frac{d x}{1 + x^{2}}

= \frac{1}{2} π μ^{- \frac{1}{2}}

F^{⁗} (μ) = - 3! \int_{0}^{\infty} \frac{d x}{{(μ + x^{2})}^{4}}} = \frac{1}{2} π \frac{d^{3} μ^{- \frac{1}{2}}}{d μ^{3}}

= \frac{1}{2} π (- \frac{1}{2}) (- \frac{3}{2}) (- \frac{5}{2}) μ^{- \frac{7}{2}}

Set

μ = 1

in both members:

\int_{0}^{\infty} \frac{d x}{{(1 + x^{2})}^{4}} = \frac{15 \frac{π}{6}}{6}

= \frac{5}{32} π

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