Неможу рішити інтеграли

Решите задачу:

$1)\quad \int \frac{\sqrt{x}\, dx}{x(x+1)}=\int }\frac{dx}{\sqrt{x}(x+1)}=[\, x=t^2,\; dx=2t\, dt\, ]=\int \frac{2t\, dt}{t(t^2+1)} =\\\\=2\int \frac{dt}{t^2+1}=2\cdot arctgt+C=2\cdot arctg\sqrt{x}+C\\\\2)\quad \int \frac{dx}{1+tgx} =[\, t=tgx,\; x=arctgt,\; dx= \frac{dt}{1+t^2} \, ]=\int \frac{dt}{(1+t^2)(1+t)} =I\\\\ \frac{1}{(1+t^2)(1+t)} = \frac{At+B}{1+t^2} + \frac{C}{1+t} = \frac{(At+B)(1+t)+C(1+t^2)}{(1+t^2)(1+t)} \\\\1=At+At^2+B+Bt+C+Ct^2\\\\t^2\; |\; A+C=0\\\\t\; |\; A+B=0\\\\t^0\; |\; B+C=1$

$A=-C\; ,\; A=-B\; ,\; \to \; \; -C=-B\; ,\; \; C=B\\\\2B=1\; \; \to \; \; B=\frac{1}{2}=C\; ,\; \; A=-\frac{1}{2}\\\\I=\int (\frac{-\frac{1}{2}t+\frac{1}{2}}{1+t^2} + \frac{\frac{1}{2}}{1+t} )dt=-\frac{1}{2}\int \frac{t\, dt}{1+t^2} + \frac{1}{2}\int \frac{dt}{1+t^2} +\frac{1}{2}\int \frac{dt}{1+t}=\\\\=-\frac{1}{4}\int \frac{d(1+t^2)}{1+t^2} +\frac{1}{2}arctgt+\frac{1}{2}ln|1+t|=\\\\=-\frac{1}{4}ln|1+t^2|+\frac{1}{2}arctgt+\frac{1}{2}ln|1+t|+C=$

$=-\frac{1}{4}ln(1+tg^2x)+\frac{1}{2}arctg(tgx)+\frac{1}{2}ln|1+tgx|+C=$

$=-\frac{1}{4}ln(1+tg^2x)+\frac{1}{2}x+\frac{1}{2}ln|1+tgx|+C$

$3)\quad \int\limits^1_0 {\frac{x^3\, dx}{3+x^4}} =\frac{1}{4}\int _0^1 \frac{4x^3\, dx}{3+x^4} =\frac{1}{4}\int _0^1 \frac{d(3+x^4)}{3+x^4} =\frac{1}{4}ln|3+x^4|\, |_0^1=\\\\=\frac{1}{4}(ln4-ln3)=\frac{1}{4}ln\frac{4}{3}$