∫(x ²+a ²) ∧-3/2 cận từ 0 → ∞

Đáp án:

 $\int^{+\infty}_0 (x^2+a^2)^\dfrac{-3}{2}dx=\dfrac{1}{a^2}$

Giải thích các bước giải:

$A=\int (x^2+a^2)^\dfrac{-3}{2}dx$

Đặt $x=atan\theta\rightarrow \theta =arctan\dfrac{x}{a}$

$\rightarrow sin\theta =\dfrac{x}{\sqrt{x^2+a^2}}$

$\rightarrow dx=\dfrac{a}{cos^2\theta}d\theta$

$\rightarrow A=\int(a^2tan^2\theta+a^2)^\dfrac{-3}{2}.\dfrac{a}{cos^2\theta}d\theta$

$\rightarrow A=\int(a^2(tan^2\theta+1))^\dfrac{-3}{2}.\dfrac{a}{cos^2\theta}d\theta$

$\rightarrow A=\int(a^2.\dfrac{1}{cos^2\theta})^\dfrac{-3}{2}.\dfrac{a}{cos^2\theta}d\theta$

$\rightarrow A=\int a^{-2}.\dfrac{1}{cos^2\theta}^\dfrac{-1}{2}d\theta$

$\rightarrow A=\int a^{-2}.cos\theta d\theta$

$\rightarrow A= a^{-2}.sin\theta$

$\rightarrow A= a^{-2}.\dfrac{x}{\sqrt{x^2+a^2}}$

$\rightarrow A|^{+\infty}_0=a^{-2}=\dfrac{1}{a^2}$