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U-substitution is the first integration technique that should be considered before pursuing the implementation of a more advanced approach. This technique, which is analogous to the chain rule of differentiation, is useful whenever a function composition can be found within the integrated.
\(\quad\displaystyle\int f’\big(g(x)\big)g'(x) = f\big(g(x)\big)+C\)
The main objective of u-substitution is to express a given integral as another integral, preferably as an integral whose antiderivative is readily known. See Integral Formulas.
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Answer the following questions by using the integration technique known as u-substitution.
By setting \(u=2x+1\) gives \(du=2dx\) and substituting these values into the integral leads to the following:
\(\normalsizeLet \(u=2x+3\), \(x = \frac{u-3}{2}\), and \(du=2dx\). Substituting these values into the integral leads to the following:
\(\normalsizeLet \(u=2x+1\), \(x = \frac{u-1}{2}\), and \(du=2dx\). Substituting these values into the integral leads to the following:
\(\normalsizeLet \(u=x^2\), and \(du=2xdx\). Substituting these values into the integral leads to the following:
\(\normalsizeLet \(u=2x^3+1\), and \(du=6x^2dx\). Substituting these values into the integral leads to the following:
\(\normalsizeLet \(u=\texttt{e}^{2x}\), and \(du=2\texttt{e}^{2x}dx\). Substituting these values into the integral leads to the following:
\(\normalsizeLet \(u=\texttt{ln}(x)\), and \(du=\frac{dx}{x}\). Substituting these values into the integral leads to the following:
\(\normalsizeLet \(u=\texttt{cos}(x)+1\), and \(du=-\texttt{sin}(x)dx\). Substituting these values into the integral leads to the following:
\(\normalsizeLet \(u=\texttt{e}^x\), and \(du=\texttt{e}^xdx\). Substituting these values into the integral leads to the following:
\(\normalsizeTransform the current integrand using algebraic manipulations:
\(\normalsize
Let \(u=\texttt{e}^x\), and \(du=\texttt{e}^xdx\). Substituting these values into the integral leads to the following:
\(\normalsizeLet \(u=\sqrt{x}\), \(du=\frac{1}{2\sqrt{x}}dx\), and \(x+1=u^2+1\). Substituting these values into the integral leads to the following:
\(\normalsizeLet \(u=x^2\), and \(du=2xdx\). Substituting these values into the integral leads to the following:
\(\normalsizeTransform the integrand using algebraic manipulations:
\(\normalsizeCheck with your tutor
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