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 $x=atan\theta\rightarrow \theta =arctan\dfrac{x}{a}$
$\rightarrow sin\theta =\dfrac{x}{\sqrt{x^2+a^2}}$
$\rightarrow dx=\dfrac{a}{cos^2\theta}d\theta$
$\rightarrow A=\int(a^2tan^2\theta+a^2)^\dfrac{-3}{2}.\dfrac{a}{cos^2\theta}d\theta$
$\rightarrow A=\int(a^2(tan^2\theta+1))^\dfrac{-3}{2}.\dfrac{a}{cos^2\theta}d\theta$
$\rightarrow A=\int(a^2.\dfrac{1}{cos^2\theta})^\dfrac{-3}{2}.\dfrac{a}{cos^2\theta}d\theta$
$\rightarrow A=\int a^{-2}.\dfrac{1}{cos^2\theta}^\dfrac{-1}{2}d\theta$
$\rightarrow A=\int a^{-2}.cos\theta d\theta$
$\rightarrow A= a^{-2}.sin\theta$
$\rightarrow A= a^{-2}.\dfrac{x}{\sqrt{x^2+a^2}}$
$\rightarrow A|^{+\infty}_0=a^{-2}=\dfrac{1}{a^2}$
