Question

Посчитать определенный интеграл. Верхний интеграл 3, нижний 0.
1/(sqrt((x^2+3)^5)). Помогите!

Answer 1

$int dfrac{dx}{sqrt{(x^2+3)^5}}=Big[; x=sqrt3tgt; ,; dx=frac{sqrt3}{cos^2t}, dt; ,; x^2+3=3, tg^2t+3=frac{3}{cos^2t}; Big]=\\\=int dfrac{sqrt3, dt}{cos^2tcdot sqrt{frac{3^5}{cos^{10}t}}}=int dfrac{cos^5t, dt}{cos^2t}=int cos^3t, dt=int cos^2tcdot cost, dt=\\\=int (1-sin^2t)cdot d(sint)=int d(sint)-int sin^2tcdot d(sint)=sint-dfrac{sin^3t}{3}+C=$

$=sin(arctgfrac{x}{sqrt3})-dfrac{sin^3(arctgfrac{x}{sqrt3})}{3}+C=dfrac{x}{sqrt{x^2+3}}-dfrac{x^3}{3, sqrt{(x^2+3)^3}}+C; ;\\\\intlimits^3_0, dfrac{dx}{sqrt{(x^2+3)^5}}=Big(dfrac{x}{sqrt{x^2+3}}-dfrac{x^3}{3, sqrt{(x^2+3)^3}}Big)Big|_0^3=dfrac{3}{sqrt{12}}-dfrac{27}{36sqrt{12}}=\\\=dfrac{sqrt3}{2}-dfrac{sqrt3}{8}=dfrac{3sqrt3}{8}$