Single Variable Calculus, Chapter 5, 5.4, Section 5.4, Problem 38

Find the integrals $\displaystyle \int^1_0 \left( 1+ x^2 \right)^3 dx$

$
\begin{equation}
\begin{aligned}
\int \left( 1+ x^2 \right)^3 dx &= \int \left( 1 + 3x^2 + 3x^4 + x^6 \right) dx\\
\\
\int \left( 1+ x^2 \right)^3 dx &= \int 1 dx + 3 \int x^2 dx + \int x^4 dx + \int x^6 dx\\
\\
\int \left( 1+ x^2 \right)^3 dx &= 1 \left( \frac{x^{0+1}}{0+1} \right) + 3 \left( \frac{x^{2+1}}{2+1} \right) + 3 \left( \frac{x^{4+1}}{4+1} \right) + \left( \frac{x^{6+1}}{6+1} \right) + C\\
\\
\int \left( 1+ x^2 \right)^3 dx &= x + \frac{\cancel{3}x^3}{\cancel{3}} + \frac{3x^5}{5} + \frac{x^7}{7} + C \\
\\
\int \left( 1+ x^2 \right)^3 dx &= x + x^3 + \frac{3x^5}{5} + \frac{x^7}{x^7} + C\\
\\
\int^1_0 \left( 1+ x^2 \right)^3 dx &= 1 + (1)^3 + \frac{3(1)^5}{5} + \frac{(1)^7}{7} + C - \left[ 0 + (0)^3 + \frac{3(0)^5}{5} + \frac{(0)^7}{5} + C \right]\\
\\
\int^1_0 \left( 1+ x^2 \right)^3 dx &= 1+1+\frac{3}{5} + \frac{1}{7} + C - 0 - 0 - 0 - 0 - C\\
\\
\int^1_0 \left( 1+ x^2 \right)^3 dx &= \frac{35+35+21+5}{35}\\
\\
\int^1_0 \left( 1+ x^2 \right)^3 dx &= \frac{96}{35}
\end{aligned}
\end{equation}
$