Question #121895

01116+9x2dx,0219+4x2dx\int_0^1 \frac{1}{\sqrt{16+9x^2}}dx, \int_0^2 \frac{1}{\sqrt{9+4x^2}}dx


1
Expert's answer
2020-06-16T18:50:14-0400

1)

01116+9x2dx=t=3x,dt=3dx,t1=0,t2=3=\int_0^1\frac{1}{\sqrt{16+9x^2}}dx=|t=3x, dt=3dx,t_1=0,t_2=3|=


1303dt42+t2=13lnt+t2+1603=ln23\frac{1}{3}\int_0^3\frac{dt}{\sqrt{4^2+t^2}}=\frac{1}{3}ln|t+\sqrt{t^2+16|}|_0^3=ln\sqrt[3]{2}

2)


0219+4x2dx=t=2x,dt=2dx,t1=0,t2=4=\int_0^2\frac{1}{\sqrt{9+4x^2}}dx=|t=2x, dt=2dx,t_1=0,t_2=4|=


1204dt32+t2=12lnt+t2+904=ln3\frac{1}{2}\int_0^4\frac{dt}{\sqrt{3^2+t^2}}=\frac{1}{2}ln|t+\sqrt{t^2+9|}|_0^4=ln\sqrt{3}


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