Question 2
Determine $\displaystyle\int\frac{e^x+1}{e^x}\,dx$
That’s it. Step through the hints below to see why.
Not quite. The sign on the second term is wrong. The antiderivative of $e^{-x}$ is $-e^{-x}$, because you divide by the $-1$ in front of the $x$. Differentiate (B) to check: $\dfrac{d}{dx}\left(x+e^{-x}\right)=1-e^{-x}$, not $1+e^{-x}$.
Not quite. (C) leaves the 1 as 1 instead of integrating it to $x$, and treats $e^{-x}$ like a constant, writing $xe^{-x}$. Differentiate (C) to check: $\dfrac{d}{dx}\left(1+xe^{-x}\right)=e^{-x}-xe^{-x}$, which isn’t $1+e^{-x}$.
Not quite. The 1 is right, but $xe^{-x}$ treats $e^{-x}$ like a constant, as if $\int k\,dx=kx$. But $e^{-x}$ changes with $x$, and its antiderivative is $-e^{-x}$. Differentiate (D) to check: you get $1+e^{-x}-xe^{-x}$, one term too many.
(A)’s derivative gives the same value as the integrand here, and at every other $x$ too. The dashed curve sits right on top of the black one, so (A) is an antiderivative.
At this $x$ both give the same value, so checking this one point would fool you. Move the guide. Everywhere else on this graph, the dashed curve misses the black one.
The curves are close here, but the values in the readout are not equal. Move the guide well away from here and the gap is clear.
The dashed curve misses the black one here, so differentiating doesn’t give back the integrand.
You can’t integrate a quotient directly. The denominator is a single term, so split the fraction.
$\dfrac{e^x+1}{e^x}=\dfrac{e^x}{e^x}+\dfrac{1}{e^x}$$=1+e^{-x}$
Integrate each term. For $e^{-x}$ the inside is linear, $-x$, so divide by its coefficient, $-1$.
$\displaystyle\int1\,dx=x$$\displaystyle\int e^{-x}\,dx=\frac{e^{-x}}{-1}=-e^{-x}$
$\displaystyle\int\left(1+e^{-x}\right)dx=x-e^{-x}+c$
Check by differentiating. The derivative of your answer must give back the integrand at every $x$.
$\dfrac{d}{dx}\left(x-e^{-x}+c\right)$$=1+e^{-x}$
The figure does this check for every option. One point isn’t enough: at $x=0$, (D)’s derivative also equals 2.
$$\begin{aligned}\int\frac{e^x+1}{e^x}\,dx&=\int\left(1+e^{-x}\right)dx\\&=x-e^{-x}+c\end{aligned}$$
Split the fraction, then integrate term by term. The antiderivative of $e^{-x}$ is $-e^{-x}$.
Only mark this done when you could do it without help. Reading the solution does not count.
Question wording is from the 2020 QCAA Mathematical Methods external assessment, © State of Queensland (QCAA) 2020, licensed under CC BY 4.0, and has been adapted. Tangent Tuition is not affiliated with the QCAA.
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