Indefinite Integral Help

**vanillachyrae** · February 18th 2013, 03:40 AM

Can you help show how to solve these??
Please kindly show the steps.

1. (x^2/3) - (x^1/4) / (x^1/2) dx
2. [x(x+2)] / (x^3 + 3x^2 - 5) dx
3. (X^3)(x^4 - 4x^2 + 4)^1/2 dx
4. [(x-1)/(x^5)]^1/2 dx
5. (x^2 + 3)^1/3 (x^5) dx
6. (x^3 + 3x^2 - 9) / (x + 3)^1/2 dx

Thanks in advance!!! :)))

**Prove It** · February 18th 2013, 04:10 AM

1. Write it as $\displaystyle \begin{align*} x^{\frac{2}{3}} - x^{-\frac{1}{4}} \end{align*}$ and integrate term by term.

2.
$\displaystyle \begin{align*} \int{\frac{x(x + 2)}{x^3 + 3x^2 - 5}\,dx} &= \int{\frac{x^2 + 2x}{x^3 + 3x^2 - 5}\,dx} \\ &= \frac{1}{3} \int{ \left( \frac{3x^2 + 6x}{x^3 + 3x^2 - 5} \right) dx } \end{align*}$

Now you can make the substitution $\displaystyle \begin{align*} u = x^3 + 3x^2 - 5 \implies du = 3x^2 + 6x \, dx \end{align*}$ and integrate.

3.
$\displaystyle \begin{align*} \int{\frac{x^3}{\sqrt{x^4 - 4x^2 + 4}}\,dx} &= \int{\frac{x^3}{\sqrt{\left( x^2 - 2 \right)^2 }} \, dx} \\ &= \int{ \frac{x^3}{ x^2 - 2 }\, dx} \\ &= \frac{1}{2} \int{ \frac{2x \cdot x^2 }{x^2 - 2} \, dx } \end{align*}$

Now make the substitution $\displaystyle \begin{align*} u = x^2 - 2 \implies du = 2x \, dx \end{align*}$ .

**earthboy** · February 18th 2013, 10:58 PM

4.

$\int \sqrt{\frac{x-1}{x^5}}dx=\int \frac{\sqrt{x-1}}{x^{2}\sqrt{x}}dx$
$=\int \frac{1}{x^2}\sqrt{\frac{x-1}{x}}dx$ now take $\frac{x-1}{x}=t^2 \implies \frac{dx}{x^2}=2tdt$
now the integral becomes $\int 2t^{2}dt = 2\frac{t^3}{3}=\frac{2}{3}(\frac{x-1}{x})^{\frac{3}{2}}$

6.

$\int \frac{x^{3}+3x^{2}-9}{\sqrt{x+3}}dx=\int\frac{x^{2}(x+3)-9}{\sqrt{x+3}}dx$
now we integrate separately
$I_{1}=\int\frac{x^{2}(x+3)}{\sqrt{x+3}}dx=\int x^{2}\sqrt{x+3}dx$ take $x+3=t^2 \implies dx=2tdt & x^2=(t^2-3)^{2}$
$I_{1}=\int (t^2-3)^{2}t^{2}dt=2\int (t^{6}-t^{4}+9t^{2})=2[\frac{t^7}{7}+\frac{6t^5}{5}+\frac{9t^3}{3}]$
substitute back the value of t to get the value in $x$
$I_{2}=\int \frac{9}{\sqrt{x+3}}$ take $x+3=t^2$ and $I_{2}$ becomes $18(x+3)^{\frac{1}{2}}$
Add $I_{1}$ and $I_{2}$ to get the result

5.

Well I tried for sometime & could not do

Spoiler:

Whats the answer ?, does it even have a elementary anti derivative?

**Prove It** · February 18th 2013, 11:21 PM

Originally Posted by vanillachyrae

Can you help show how to solve these??
Please kindly show the steps.

1. (x^2/3) - (x^1/4) / (x^1/2) dx
2. [x(x+2)] / (x^3 + 3x^2 - 5) dx
3. (X^3)(x^4 - 4x^2 + 4)^1/2 dx
4. [(x-1)/(x^5)]^1/2 dx
5. (x^2 + 3)^1/3 (x^5) dx
6. (x^3 + 3x^2 - 9) / (x + 3)^1/2 dx

Thanks in advance!!!

))

5.
$\displaystyle \begin{align*} \int{x^5 \left( x^2 + 3 \right)^{\frac{1}{3}} \, dx} &= \frac{1}{2} \int{ 2x \cdot x^4 \left( x^2 + 3 \right)^{\frac{1}{3}}\,dx } \end{align*}$

Now let $\displaystyle \begin{align*} u = x^2 + 3 \implies du = 2x \, dx \end{align*}$ and the integral becomes

$\displaystyle \begin{align*} \frac{1}{2}\int{ 2x \cdot x^4 \left( x^2 + 3 \right)^{\frac{1}{3}}\, dx} &= \frac{1}{2} \int{ \left( u - 3 \right)^2 \, u^{\frac{1}{3}}\, du } \\ &= \frac{1}{2} \int{ \left( u^2 - 6u + 9 \right) u^{\frac{1}{3}}\, du } \\ &= \frac{1}{2} \int{ u^{\frac{7}{3}} - 6u^{\frac{4}{3}} + 9u^{\frac{1}{3}}\, du } \\ &= \frac{1}{2} \left( \frac{3}{10}u^{\frac{10}{3}} - \frac{18}{7}u^{\frac{7}{3}} + \frac{27}{4} u^{\frac{4}{3}} \right) + C \\ &= \frac{3}{20} \left( x^2 + 3 \right)^{\frac{10}{3}} - \frac{9}{7} \left( x^2 + 3 \right)^{\frac{7}{3}} + \frac{27}{8} \left( x^2 + 3 \right)^{\frac{4}{3}} + C \end{align*}$

**ibdutt** · February 19th 2013, 02:36 AM

Q No 5 Attached

**Prove It** · February 19th 2013, 04:03 AM

Originally Posted by ibdutt

Q No 5 Attached

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Is this not exactly what I did in the post directly above?

**earthboy** · February 19th 2013, 05:08 AM

Hmm..seems like I got the whole question 5 wrong...

@ vanillachyrae : please learn to write in latex(its not at all hard)

@ Prove it: You were great and quick as usual

**Kmath** · February 19th 2013, 05:18 AM

Originally Posted by Prove It

5.
$\displaystyle \begin{align*} \int{x^5 \left( x^2 + 3 \right)^{\frac{1}{3}} \, dx} &= \frac{1}{2} \int{ 2x \cdot x^4 \left( x^2 + 3 \right)^{\frac{1}{3}}\,dx } \end{align*}$

Now let $\displaystyle \begin{align*} u = x^2 + 3 \implies du = 2x \, dx \end{align*}$ and the integral becomes

$\displaystyle \begin{align*} \frac{1}{2}\int{ 2x \cdot x^4 \left( x^2 + 3 \right)^{\frac{1}{3}}\, dx} &= \frac{1}{2} \int{ \left( u - 3 \right)^2 \, u^{\frac{1}{3}}\, du } \\ &= \frac{1}{2} \int{ \left( u^2 - 6u + 9 \right) u^{\frac{1}{3}}\, du } \\ &= \frac{1}{2} \int{ u^{\frac{7}{3}} - 6u^{\frac{4}{3}} + 9u^{\frac{1}{3}}\, du } \\ &= \frac{1}{2} \left( \frac{3}{10}u^{\frac{10}{3}} - \frac{18}{7}u^{\frac{7}{3}} + \frac{27}{4} u^{\frac{4}{3}} \right) + C \\ &= \frac{3}{20} \left( x^2 + 3 \right)^{\frac{10}{3}} - \frac{9}{7} \left( x^2 + 3 \right)^{\frac{7}{3}} + \frac{27}{8} \left( x^2 + 3 \right)^{\frac{4}{3}} + C \end{align*}$

Also

$5.\displaystyle \begin{align*} \int{x^5 \left( x^2 + 3 \right)^{\frac{1}{3}} \, dx} \end{align*}$
now let
$\displaystyle \begin{align*} t = x^2 \implies dt = 2x \, dx \end{align*}$ and the integral becomes

$\frac{1}{2}\int{ t^2(t+3)^{\frac{1}{3}}dt}$

by parts
let $u=t^2$ and $dv=(t+3)^{\frac{1}{3}}dt$
and once again with integration by parts

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