What is $\int x^{3} \cdot \sqrt{x^{2} + 4} d x$ $int x^3 * sqrt(x^2 + 4)dx$ ?

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What is $\int x^{3} \cdot \sqrt{x^{2} + 4} d x$ $int x^3 * sqrt(x^2 + 4)dx$ ?

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Answer 1 · 2015-12-23T01:51:36

Use the substitution $u = x^{2} + 4$ $u = x^2+4$ to find

$\int x^{3} \sqrt{x^{2} + 4} d x = \frac{1}{5} {(x^{2} + 4)}^{\frac{5}{2}} - \frac{4}{3} {(x^{2} + 4)}^{\frac{3}{2}} + C$ $intx^3sqrt(x^2+4)dx= 1/5(x^2+4)^(5/2) - 4/3(x^2+4)^(3/2)+C$

Explanation:

For this problem we will use:

$\int x^{n} d x = \frac{x^{n + 1}}{n + 1} + C$ $intx^ndx = x^(n+1)/(n+1) + C$
$\int (f (x) + g (x)) d x = \int f (x) d x + \int g (x) d x$ $int(f(x) + g(x))dx = intf(x)dx + intg(x)dx$
Integration by Substitution

We will proceed using the substitution $u = x^{2} + 4$ $u = x^2 + 4$ .
Then $d u = 2 x d x$ $du = 2xdx$ and we have

$\int x^{3} \sqrt{x^{2} + 4} d x = \frac{1}{2} \int ((x^{2} + 4) - 4) \sqrt{x^{2} + 4} \cdot 2 x d x$ $intx^3sqrt(x^2+4)dx = 1/2int((x^2+4)-4)sqrt(x^2+4)*2xdx$

$= \frac{1}{2} \int (u - 4) \sqrt{u} d u$ $= 1/2int(u-4)sqrt(u)du$

$= \frac{1}{2} \int (u^{\frac{3}{2}} - 4 u^{\frac{1}{2}}) d u$ $=1/2int(u^(3/2)-4u^(1/2))du$

$= \frac{1}{2} \int u^{\frac{3}{2}} d u - 2 \int u^{\frac{1}{2}} d u$ $=1/2intu^(3/2)du - 2intu^(1/2)du$

$= \frac{1}{5} u^{\frac{5}{2}} - \frac{4}{3} u^{\frac{3}{2}} + C$ $=1/5u^(5/2) - 4/3u^(3/2)+C$

$= \frac{1}{5} {(x^{2} + 4)}^{\frac{5}{2}} - \frac{4}{3} {(x^{2} + 4)}^{\frac{3}{2}} + C$ $= 1/5(x^2+4)^(5/2) - 4/3(x^2+4)^(3/2)+C$

Answer 2 · 2015-12-23T02:02:23

Use $u$ $u$ substitution and then solve.

Explanation:

$\int x^{3} \cdot \sqrt{x^{2} + 4} d x$ $int x^3 * sqrt(x^2 + 4) dx$

Let $u = x^{2} + 4$ $u = x^2 + 4$

Differentiate with respect to $x$ $x$

$d u = 2 x d x$ $du = 2x dx$
$\frac{d u}{2} = x d x$ $(du)/2 = x dx$

Also we have $u = x^{2} + 4$ $u=x^2+4$ therefore $x^{2} = u - 4$ $x^2 = u-4$

$\int x^{3} \cdot \sqrt{x^{2} + 4} d x$ $int x^3 * sqrt(x^2 + 4) dx$

Rewriting our integral as
$\int x^{2} \cdot \sqrt{x^{2} + 4} x d x$ $int x^2 * sqrt(x^2 + 4) x dx$

Substituting for $x^{2}, x^{2} + 4$ $x^2, x^2+4$ and $x d x$ $x dx$

$\int (u - 4) \sqrt{u} \frac{d u}{2}$ $int (u-4) sqrt(u) (du)/2$
$\frac{1}{2} \int (u - 4) \sqrt{u} d u$ $1/2 int (u-4) sqrt(u) du$
$\frac{1}{2} \int (u^{\frac{3}{2}} - 4 u^{\frac{1}{2}}) d u$ $1/2 int (u^(3/2) - 4u^(1/2)) du$ Note $\sqrt{u} = u^{\frac{1}{2}}$ $sqrt(u) = u^(1/2)$
$\frac{1}{2} {\int u^{\frac{3}{2}} d u - 4 \int u^{\frac{1}{2}} d u}$ $1/2 { int u^(3/2) du - 4 int u^(1/2) du }$

We apply power rule $\int x^{n} d x = \frac{x^{n + 1}}{n + 1} + C$ $int x^n dx = x^(n+1)/(n+1) + C$ where $x$ $x$ not equal to $1$ $1$

$\frac{1}{2} {\frac{u^{\frac{3}{2} + 1}}{\frac{3}{2} + 1} - 4 \frac{u^{\frac{1}{2} + 1}}{\frac{1}{2} + 1}} + C$ $1/2 { u^(3/2 + 1)/(3/2+1) - 4 u^(1/2 + 1)/(1/2+1)} + C$
$\frac{1}{2} {\frac{u^{\frac{5}{2}}}{\frac{5}{2}} - 4 \frac{u^{\frac{3}{2}}}{\frac{3}{2}}} + C$ $1/2{ u^(5/2)/(5/2) - 4 u^(3/2)/(3/2)} + C$
$\frac{1}{2} {(\frac{2}{5}) u^{\frac{5}{2}} - 4 (\frac{2}{3}) u^{\frac{3}{2}}} + C$ $1/2 {(2/5)u^(5/2) - 4 (2/3) u^(3/2)}+C$
$\frac{1}{5} u^{\frac{5}{2}} - \frac{4}{3} u^{\frac{3}{2}} + C$ $1/5 u^(5/2) - 4/3 u^(3/2) + C$
$\frac{1}{5} {(x^{2} + 4)}^{\frac{5}{2}} - \frac{4}{3} {(x^{2} + 4)}^{\frac{3}{2}} + C$ $1/5 (x^2+4)^(5/2) - 4/3 (x^2+4)^(3/2) +C$ Answer .

It can be further simplified if needed.

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2 Answers

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