Question

Посчитать определенный интеграл. Верхний интеграл 3, нижний

Answer 1

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

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