Can the Chain Rule Help Me Integrate This Tricky Function?

[karush]

$$\tiny\text{Whitman 8.7.15 Chain Rule} $$
$$\displaystyle
I=\int \frac{\sec^2\left({t}\right)}{\left(1+\tan\left({t}\right)\right)^2}\ d{t}
=\frac{-1}{2\left(1+\tan\left({t}\right)\right)}
+ C$$
$\begin{align}\displaystyle
u& = \tan\left({t}\right)&
du&= \sec^2 \left({t}\right)\ d{t} \\
\end{align}$

$\text{so.. chain rule} $

$
\displaystyle
\int {x}^{n} \ dx = \frac{{x}^{x+1}}{n+1}+C
$

$\text{rewrite and integrate} $
$$I=\int \frac{1}{\left(1+u\right)^3} d{u}
=\frac{-1}{2\left(1+\tan\left({t}\right)\right)^2}
+ C
$$

$\tiny\text
{from Surf the Nations math study group}$

 

[Prove It]

$$\tiny\text{Whitman 8.7.15 Chain Rule} $$
$$\displaystyle
I=\int \frac{\sec^2\left({t}\right)}{\left(1+\tan\left({t}\right)\right)^2}\ d{t}
=\frac{-1}{2\left(1+\tan\left({t}\right)\right)}
+ C$$
$\begin{align}\displaystyle
u& = \tan\left({t}\right)&
du&= \sec^2 \left({t}\right)\ d{t} \\
\end{align}$

$\text{so.. chain rule} $

$
\displaystyle
\int {x}^{n} \ dx = \frac{{x}^{x+1}}{n+1}+C
$

$\text{rewrite and integrate} $
$$I=\int \frac{1}{\left(1+u\right)^3} d{u}
=\frac{-1}{2\left(1+\tan\left({t}\right)\right)^2}
+ C
$$

$\tiny\text
{from Surf the Nations math study group}$

You somehow changed the integrand into the reciprocal of a cubic when it was originally the reciprocal of a quadratic...

 

[karush]

The given should of been a cube. Thanks for the catch. Hope the rest is OK.