Ejercicios de Integrales Resueltas

17-1A) $\displaystyle \int_{-3}^3 \sqrt{3+|x|} \, dx = $ $\displaystyle \int_{-3}^0 \sqrt{3-x} \, dx + \int_{0}^3 \sqrt{3+x} \, dx =$ $[t=3-x], [t=3+x] =$ $\displaystyle \int_{6}^3 \sqrt{t} (-1) \,dt + \int_{3}^6 \sqrt{t} \, dt=$ $ 8 \sqrt{6} – 4 \sqrt{3}$

17-2B) $\displaystyle \int \dfrac{x^2}{\sqrt{(x^3+1)^3}} \, dx = $ $[t = x^3 + 1 ] =$ $\displaystyle \dfrac{1}{3} \int t^{-3}{2} \, dt = $ $\displaystyle \dfrac{-2}{3\sqrt{(x^3+1)^3}} + C$

17-1B) $\displaystyle \int_0^{\sqrt{3}} x \, arc\,tan(x) \, dx = $ $[u = arc\tan (x) ; dv = x] =$ $\displaystyle \left[\dfrac{x^2}{2} \, arc\, tan (x) \right]_0^{\sqrt{3}} – \int_0^{\sqrt{3}} \dfrac{x^2}{2} \dfrac{1}{1+x^2} \, dx = $ $ \displaystyle \left( \sqrt{3} \, \dfrac{\pi}{3} \, – \, 0 \right) \, – \dfrac{1}{2} \int_0^{\sqrt{3}} 1 – \dfrac{1}{1+x^2} = $ $\displaystyle \dfrac{\sqrt{3}}{3}\pi – \dfrac{1}{2} \left[ x – arctan(x) \right]_0^{\sqrt{3}} = $ $\displaystyle \dfrac{\sqrt{3}}{3}\pi – \dfrac{\sqrt{3}}{2} + \dfrac{\pi}{6} $

17-2A) $\displaystyle \int (x+2) \, ln(x) \, dx $ $[u =ln (x) ; dv = x+2] =$ $\displaystyle \left( \dfrac{x^2}{2} + 2x \right) \, lnx – \int \dfrac{1}{x} \cdot \left(\dfrac{x^2}{2} +2 \right) \, dx =$ $\displaystyle \left( \dfrac{x^2}{2} + 2x \right) \, lnx – \dfrac{x^2}{4} – 2x + C$

17-JB) $\displaystyle \int_1^{16} \dfrac{dx}{\sqrt{x} + \sqrt[4]{x}} = $ $[t=\sqrt[4]{x}] =$ $\displaystyle \int_1^{2} \dfrac{4t^3}{t^2+t} dt = $ $\displaystyle \int_1^{2} 4t – 4 + \dfrac{4}{t+1} dt = 2 + 4 ln\dfrac{3}{2}$

17-3A) $\displaystyle \int_0^{8} \dfrac{dx}{2 + \sqrt{x+1}} =$ $[t = 2 + \sqrt{x+1} ] =$ $\displaystyle 2 \int_3^{5} \dfrac{t – 2}{t} \, dt = $ $\displaystyle 2 \left[ t – 2 ln\,t \right]_3^5 = 4 + 4 ln \dfrac{3}{5}$