Треба обчислити інтеграли

[latex]1)\quad \int \frac{x\, dx}{\sqrt{x^4+1}}=\frac{1}{2}\int \frac{2x\, dx}{\sqrt{(x^2)^2+1}}=\frac{1}{2}\int \frac{dt}{\sqrt{t^2+1}}=\\\\=\frac{1}{2}\cdot ln|t+\sqrt{t^2+1}+C=\frac{1}{2}\cdot ln|x^2+\sqrt{x^4+1}|+C\\\\2)\quad \int arcsinx\, dx=[u=arcsinx,\; du=\frac{dx}{\sqrt{1-x^2}},\; v=x]=\\\\=uv-\int v\, du=x\cdot arcsinx-\int \frac{x\, dx}{\sqrt{1-x^2}}=\\\\=x\cdot arcsinx-\frac{1}{2}\int \frac{-d(1-x^2)}{\sqrt{1-x^2}}=x\cdot arcsinx+\frac{1}{2}\cdot 2\sqrt{1-x^2}+C[/latex] [latex]3)\quad \int \frac{x\, dx}{x^2+4x+5} =\int \frac{x\, dx}{(x+2)^2+1} =[t=x+2,\; dt=dx]=\\\\=\int \frac{(t-2)dt}{t^2+1} =\frac{1}{2}\int \frac{2t\, dt}{t^2+1} -2\int \frac{dt}{t^2+1} =\\\\=\frac{1}{2}\cdot ln|t^2+1|-2arctgt+C=\\\\=\frac{1}{2}\cdot ln|x^2+4x+5|-2arctg(x+2)+C\\\\4)\quad \int\limits^{63}_0 {\frac{x\, dx}{\sqrt[3]{x+1}}} \, dx =[t^3=x+1,x=t^3-1,\; dx=3t^2\, dt,\\\\t_1=4,t_2=1\, ]=\int _1^4\frac{t^3-1}{t}dt=\int _1^4(t^2-\frac{1}{t})dt=[/latex] [latex]=(\frac{t^3}{3}-ln|t|)|_1^4=\frac{64}{3}-ln4-(\frac{1}{3}-ln1)=[/latex] [latex]=21-ln4[/latex]