Неопределен интеграл

[Гост]

Някой може ли да помогне с решението на този интеграл

[ammornil]

[tex]\int \left(\frac{1}{1+9x^2}+\frac{x}{5x+1}+\frac{tgx}{cos^2x}\right)dx=? \\
\vspace{10} \\
\cyr{DM}: \left|1+9x^2 \ne 0 \\ 5x+1 \ne 0 \\ cos^2x \ne 0 \right, \hspace{4} \Rightarrow \left| \forall x \in R \\ x \ne -\frac{1}{5} \\ x \ne \pm k.\pi \hspace{4} (k \in N) \right, \\
\vspace{10} \\
\int \left(\frac{1}{1+9x^2}+\frac{x}{5x+1}+\frac{tgx}{cos^2x}\right)dx=\int \frac{dx}{1+(3x)^2}+\int \frac{xdx}{5x+1}+\int \frac{sinx}{cos^3 x}dx=\\ \vspace{5} \\
=arctg(3x)+C_1+\frac{1}{5}.\int \frac{5x+1-1}{5x+1}dx-\int \frac{dcosx}{cos^3 x}= \\ \vspace{5} \\
=arctg(3x)+C_1+\frac{1}{5}\int dx-\frac{1}{5}\int \frac{dx}{5x+1}+
\frac{1}{2.cos^2 x}+C_4= \\ \vspace{5} \\
=arctg(3x)+C_1+\frac{1}{5}.x+C_2-\frac{1}{5}\int \frac{\frac{1}{5}.d(5x+1)}{5x+1}+\frac{1}{2.cos^2 x}+C_4= \\ \vspace{5} \\
=arctg(3x)+C_1+\frac{1}{5}.x+C_2-\frac{1}{25}ln(5x+1)+C_3+\frac{1}{2.cos^2 x}+C_4= \\ \vspace{5} \\
=arctg(3x)+\frac{1}{5}.x-\frac{1}{25}ln(5x+1)+\frac{1}{2.cos^2 x}+C[/tex]

[ammornil]

Имам разменени знаци по-горе...
[tex]\int \left(\frac{1}{1+9x^2}+\frac{x}{5x+1}+\frac{tgx}{cos^2x}\right)dx=? \\
\vspace{10} \\
\cyr{DM}: \left|1+9x^2 \ne 0 \\ 5x+1 \ne 0 \\ cos^2x \ne 0 \right, \hspace{4} \Rightarrow \left| \forall x \in R \\ x \ne -\frac{1}{5} \\ x \ne \pm k.\pi \hspace{4} (k \in N) \right, \\
\vspace{10} \\
\int \left(\frac{1}{1+9x^2}+\frac{x}{5x+1}+\frac{tgx}{cos^2x}\right)dx=\int \frac{dx}{1+(3x)^2}+\int \frac{xdx}{5x+1}+\int \frac{sinx}{cos^3 x}dx=\\ \vspace{5} \\
=arctg(3x)+C_1+\frac{1}{5}.\int \frac{5x+1-1}{5x+1}dx-\int \frac{dcosx}{cos^3 x}= \\ \vspace{5} \\
=arctg(3x)+C_1-\frac{1}{5}\int dx+\frac{1}{5}\int \frac{dx}{5x+1}+
\frac{1}{2.cos^2 x}+C_4= \\ \vspace{5} \\
=arctg(3x)+C_1-\frac{1}{5}.x+C_2-+\frac{1}{5}\int \frac{\frac{1}{5}.d(5x+1)}{5x+1}+\frac{1}{2.cos^2 x}+C_4= \\ \vspace{5} \\
=arctg(3x)+C_1-\frac{1}{5}.x+C_2+\frac{1}{25}ln(5x+1)+C_3+\frac{1}{2.cos^2 x}+C_4= \\ \vspace{5} \\
=arctg(3x)-\frac{1}{5}.x+\frac{1}{25}ln(5x+1)+\frac{1}{2.cos^2 x}+C[/tex]

[Гост]

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