The integral of the expression \(x^n\) where \(n \neq -1\) is:

\[ \int x^ndx = \frac{x^{n+1}}{n+1} + C \]

The integral of \(x^{-1}\) is:

\[ \int \frac{1}{x}\,dx = \ln \left| x \right| + C \]

Note the absolute value inside the natural log.

The integral of the exponential function \( e^x \) is:

\[ \int e^x \, dx = e^x + C \]

## Find a: \( \displaystyle \int_0^{\ln a} e^x \, dx = 100 \)

Since \( \int e^x \, dx = e^x + C \), we see that:

\[ \begin{align} \int_0^{\ln a} e^x \, dx &= 100 \\ \left[ e^x \right]_0^{\ln a} &= 100 \\ e^{\ln a}-e^0 &= 100 \\ a &= 101 _\square \end{align} \]

## Evaluate: \( \displaystyle \int_0^1 7x^6 + 42 + \frac{2}{x^3} \, dx \).

We can rewrite this problem so that all terms are in the form \(x^n\):

\[ \displaystyle \int_0^1 7x^6 + 42x^0 + 2x^{-3} \, dx \]

Now, let's integrate, using the rules above:

\[ \begin{align} \int_0^1 7x^6 + 42 + 2x^{-3} \, dx &= \int_0^1 7x^6 dx + \int_0^1 42 x^0 \, dx + \int_0^1 2x^{-3}\, dx \\

&= \left[ 7\left(\frac{x^{6+1}}{6+1} \right ) + 42x^1 + 2 \left( \frac{x^{-3+1}}{-3+1} \right ) \right]_0^1 \\

&=\left. x^7 + 42x -x^{-2} \right|_0^1 \\ &= \left(1^7 +42(1)-1\right)-\left(0\right) \\ &= 42 \, _\square \end{align} \]

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