# How do you use substitution to integrate x^2sqrt(4x+9)dx?

Jun 16, 2015

The answer can be written as $\int {x}^{2} \sqrt{4 x + 9} \setminus \mathrm{dx} = \setminus \frac{10 {x}^{2} - 18 x + 27}{140} {\left(4 x + 9\right)}^{\frac{3}{2}} + C$

#### Explanation:

Use the substitution $u = 4 x + 9$ so that $\mathrm{du} = 4 \mathrm{dx}$ and $x = \frac{u - 9}{4} = \frac{1}{4} u - \frac{9}{4}$. Then

$\int {x}^{2} \sqrt{4 x + 9} \setminus \mathrm{dx} = \frac{1}{4} \int {\left(\frac{u - 9}{4}\right)}^{2} \sqrt{u} \setminus \mathrm{du}$

$= \frac{1}{64} \int \left({u}^{2} - 18 u + 81\right) {u}^{\frac{1}{2}} \setminus \mathrm{du}$

$= \frac{1}{64} \int \left({u}^{\frac{5}{2}} - 18 {u}^{\frac{3}{2}} + 81 {u}^{\frac{1}{2}}\right) \setminus \mathrm{du}$

$= \frac{1}{64} \cdot \frac{2}{7} {u}^{\frac{7}{2}} - \frac{9}{32} \cdot \frac{2}{5} {u}^{\frac{5}{2}} + \frac{81}{64} \cdot \frac{2}{3} {u}^{\frac{3}{2}} + C$

$= \left(\frac{1}{224} {u}^{2} - \frac{9}{80} u + \frac{27}{32}\right) {u}^{\frac{3}{2}} + C$

Since ${u}^{2} = {\left(4 x + 9\right)}^{2} = 16 {x}^{2} + 72 x + 81$, we can write this as

$= \left(\frac{1}{14} {x}^{2} + \frac{9}{28} x + \frac{81}{224} - \frac{9}{20} x - \frac{81}{80} + \frac{27}{32}\right) {\left(4 x + 9\right)}^{\frac{3}{2}} + C$

$= \setminus \frac{10 {x}^{2} - 18 x + 27}{140} {\left(4 x + 9\right)}^{\frac{3}{2}} + C$

Jun 17, 2015

You can rewrite this carefully.

$= \int {x}^{2} \sqrt{{\left(2 \sqrt{x}\right)}^{2} + {3}^{2}} \mathrm{dx}$

Now it looks like $\sqrt{{a}^{2} + {x}^{2}}$, which resembles $\sqrt{1 + {\tan}^{2} \theta}$.

Let:
$2 \sqrt{x} = 3 \tan \theta$
$x = \frac{9 {\tan}^{2} \theta}{4}$
${x}^{2} = \frac{81 {\tan}^{4} \theta}{16}$
$\sqrt{{\left(2 \sqrt{x}\right)}^{2} + {3}^{2}} = \sqrt{9 {\tan}^{2} \theta + 9} = 3 \sec \theta$
$\mathrm{dx} = \frac{9}{4} \cdot 2 \tan \theta {\sec}^{2} \theta d \theta = \frac{9}{2} \tan \theta {\sec}^{2} \theta d \theta$

$\implies \int \frac{81 {\tan}^{4} \theta}{16} 3 \sec \theta \frac{9}{2} \tan \theta {\sec}^{2} \theta d \theta$

$= \frac{2187}{32} \int {\tan}^{5} \theta {\sec}^{3} \theta d \theta$

Notice how that constant is irreducible... Ugh. It looks a bit crazy, but it's just large numbers. Using some identities:
$= \frac{2187}{32} \int {\left({\sec}^{2} \theta - 1\right)}^{2} {\sec}^{2} \theta \sec \theta \tan \theta d \theta$

And some u-substitution. Let:
$u = \sec \theta$
$\mathrm{du} = \sec \theta \tan \theta d \theta$

$= \frac{2187}{32} \int {\left({u}^{2} - 1\right)}^{2} {u}^{2} \mathrm{du}$

$= \frac{2187}{32} \int \left({u}^{4} - 2 {u}^{2} + 1\right) {u}^{2} \mathrm{du}$

$= \frac{2187}{32} \int {u}^{6} - 2 {u}^{4} + {u}^{2} \mathrm{du}$

$= \frac{2187}{32} \left({u}^{7} / 7 - \frac{2}{5} {u}^{5} + {u}^{3} / 3\right)$

$= \frac{2187}{32} \left({\sec}^{7} \frac{\theta}{7} - \frac{2}{5} {\sec}^{5} \theta + {\sec}^{3} \frac{\theta}{3}\right)$

Draw a right triangle if you want. Since $\tan \theta = \frac{2 \sqrt{x}}{3}$, and $\sqrt{{\left(2 \sqrt{x}\right)}^{2} + {3}^{2}} = \sqrt{4 x + 9}$, $\sec \theta = \frac{\sqrt{4 x + 9}}{3}$:

$= \frac{2187}{32} \left({\left(4 x + 9\right)}^{\frac{7}{2}} / \left(7 \cdot {3}^{7}\right) - \frac{2}{5 \cdot {3}^{5}} {\left(4 x + 9\right)}^{\frac{5}{2}} + {\left(4 x + 9\right)}^{\frac{3}{2}} / \left(3 \cdot {3}^{3}\right)\right)$

$= \frac{2187}{32} \left({\left(4 x + 9\right)}^{\frac{7}{2}} / \left(15309\right) - \frac{2}{1215} {\left(4 x + 9\right)}^{\frac{5}{2}} + {\left(4 x + 9\right)}^{\frac{3}{2}} / \left(81\right)\right)$

$= \frac{1}{32} \left({\left(4 x + 9\right)}^{\frac{7}{2}} / \left(7\right) - \frac{18}{5} {\left(4 x + 9\right)}^{\frac{5}{2}} + 27 {\left(4 x + 9\right)}^{\frac{3}{2}}\right)$

$= \frac{1}{32} {\left(4 x + 9\right)}^{\frac{3}{2}} \left[{\left(4 x + 9\right)}^{2} / 7 - \frac{18}{5} \left(4 x + 9\right) + 27\right] + C$

Now let's figure out how to make this look nicer. Get some common denominators, subtract in the numerator, and factor out $\frac{8}{35}$:
$= \frac{1}{32} {\left(4 x + 9\right)}^{\frac{3}{2}} \left[\frac{16 {x}^{2} + 72 x + 81}{7} - \frac{72}{5} x - \frac{162}{5} + 27\right]$

$= \frac{1}{32} {\left(4 x + 9\right)}^{\frac{3}{2}} \left[\frac{80 {x}^{2} + 360 x + 405 - 504 x - 1134 + 945}{35}\right]$

$= \frac{1}{32} {\left(4 x + 9\right)}^{\frac{3}{2}} \left[\frac{80 {x}^{2} - 144 x + 216}{35}\right]$

$= \frac{1 \cdot 8}{32 \cdot 35} {\left(4 x + 9\right)}^{\frac{3}{2}} \left[10 {x}^{2} - 18 x + 27\right]$

$= \frac{1}{140} {\left(4 x + 9\right)}^{\text{3/2}} \left[10 {x}^{2} - 18 x + 27\right] + C$