Помогите решить 6.3 и 6.14 даю 30 баллов. Created by SBT. algebra-ru

Помогите решить 6.3 и 6.14 даю 30 баллов...

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подкожного пожалуйста ,и спасибо за оценку

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$6.3)\; \; \int \frac{2x+1}{5x^2+1} dx=\int \frac{2x\cdot dx}{5x^2+1} +\int \frac{dx}{5x^2+1} =\\\\=\frac{1}{5}\cdot \int \frac{10x\cdot dx}{5x^2+1}+\frac{1}{5}\int \frac{dx}{x^2+\frac{1}{5}} =[\; t=5x^2+1\; ,\; dt=10x\, dx\; ]=\\\\= \frac{1}{5}\cdot \int \frac{dt}{t}+ \frac{1}{5}\cdot \frac{1}{\sqrt{\frac{1}{5}}} arctg \frac{x}{\sqrt{\frac{1}{5}}} +C= \frac{1}{5} \cdot ln|t|+ \frac{\sqrt5}{5} \cdot arctg(\sqrt5x)+C=\\\\= \frac{1}{5}\cdot ln|5x^2+1|+\frac{1}{\sqrt5}\cdot arctg(\sqrt5x)+C$

$6.14)\; \; \int \frac{5x-2}{x^2+9} dx=5\cdot \int \frac{x\cdot dx}{x^2+9} -2\cdot \int \frac{dx}{x^2+9} =\\\\=\frac{5}{2}\int \frac{2x\cdot dx}{x^2+9} -2\cdot \frac{1}{3}\cdot arctg\frac{x}{3}=\frac{5}{2}\int \frac{d(x^2+9)}{x^2+9}-\frac{2}{3}\cdot arctg\frac{x}{3}=\\\\=[\; t=x^2+9\; ,\; dt=2x\, dx\; ,\; \int \frac{dt}{t}=ln|t|+C\; ]=\\\\= \frac{5}{2} \cdot ln|x^2+9|- \frac{2}{3}\cdot arctg\frac{x}{3}+C$

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