Задание 10.2

Question

Найдите общий вид первообразной функции:
\begin{enumerate}
\item $f(x)=x+3;$
\item $f(x) = x^2+4x-1;$
\item $f(x) = \frac{x^3+x}{x^2+1};$
\item $f(x) = \frac{1}{2}e^x+2^x\ln 2;$
\item $f(x) = \frac{9}{\cos^2 x}-3\sin x;$
\item $f(x) = 5\sqrt[4]{x}-\frac{3}{x};$
\item $f(x) = 6x^2-\frac{2}{x^2};$
\item $f(x) = \frac{9}{x^10}+\frac{8}{x^9}.$
\end{enumerate}

Gimli · Accepted Answer

\begin{enumerate}
\item $$\int (x+3)dx = \int xdx + \int 3dx = \frac{1}{2}x^2+3x+C.$$
\item $$\int (x^2+4x-1)dx = \int x^2dx + \int 4xdx +\int (-1)dx = \frac{1}{3}x^3+2x^2-x+C.$$
\item $$\int \frac{x^3+x}{x^2+1} dx = \int \frac{x(x^2+1)}{x^2+1} dx=\int xdx=\frac{1}{2}x^2+C.$$
\item $$\int \left(\frac{1}{2}e^x+2^x\ln 2ight)dx = \int \frac{1}{2} e^x dx+ \int 2^x \ln 2 dx = \frac{1}{2}e^x+2^x+C.$$
\item $$\int \left(\frac{9}{\cos^2 x}-3\sin xight)dx = \int \frac{9}{\cos^2 x}dx + \int -3\sin x dx = 9\operatorname{	g} x+3\cos x+C.$$
\item $$\int \left(5\sqrt[4]{x} - \frac{3}{x}ight)dx = \int 5\sqrt[4]{x}dx + \int -\frac{3}{x} dx = 4\sqrt[4]{x^5}-3\ln x + C.$$
\item $$\int \left(6x^2-\frac{2}{x^2}ight)dx=2x^3+\frac{2}{x}+C.$$
\item $$\int \left(\frac{9}{x^{10}}+\frac{8}{x^9}ight)dx= (-x)^{-9}-(-x)^{-8}+C.$$
\end{enumerate}

Задание 10.2

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Задание 10.2

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