Question

Use the specified substitution to find or evaluate the integral. $$ \begin{array}{l} \int \sqrt{e^{t}-3} d t \\ u=\sqrt{e^{t}-3} \end{array} $$

Short answer

The integral of \(\sqrt{e^{t}-3}\) with respect to \(t\), given the substitution \(u=\sqrt{e^{t}-3}\), is \(1/2 * ln|\sqrt{e^{t}-3}| + C\).

Step-by-step solution

Find the substitution derivative

We have that \(u=\sqrt{e^{t}-3}\). Squaring both sides yields \(u^2=e^{t}-3\). When we add 3 to both sides and take the logarithm (base \(e\)) of both sides, we get \(e^{t}=u^2+3\). This lets us find the derivative \(dt\) in terms of \(du\). The derivative of \(e^{t}\) with respect to \(t\) is \(e^{t}\), which would yield \(dt = du/(2u)\).

Express the integral in terms of u

Substitute \(u\) and \(dt\) into the integral. The integral transforms from \(\int \sqrt{e^{t}-3} dt\) to \(\int u * du/(2u)\).

Simplify and Compute Integral

The \(u\) terms cancel out, yielding \(\int 1/(2u) du\). To find the integral, integrate with respect to \(u\) which yields \(1/2 * ln|u| + C\).

Substitute back for t

Substitute back for \(t\) by replacing \(u\) by \(\sqrt{e^{t}-3}\) in the expression obtained. This gives the final result as \(1/2 * ln|\sqrt{e^{t}-3}| + C\).