1. ## Stuck with Integration(Logarithms)

This was our quiz last week which i failed(0/5), hah,, Can you show me how do I integrate this? What methods did my classmates used? Thanks.

$\int \frac{x{dx}}{(x^2-1)ln(1-x^2)}$

$\int\frac{sin4xdx}{1+sin^22x}$

$\int\frac{\sqrt x dx}{1-\sqrt{x^3}}$

$\int\frac{(2x-3) dx}{(x-2)^2}$

$\int\frac{(3-4x)x dx}{\sqrt[3]{1-x}}$

2. Originally Posted by honestliar
This was our quiz last week which i failed(0/5), hah,, Can you show me how do I integrate this? What methods did my classmates used? Thanks.

$\int \frac{x{dx}}{(x^2-1)ln(1-x^2)}$
$\int {\frac{x}{{\left( {{x^2} - 1} \right)\ln \left( {1 - {x^2}} \right)}}dx} = \frac{1}{2}\int {\frac{{ - 2x}}{{\left( {1 - {x^2}} \right)\ln \left( {1 - {x^2}} \right)}}dx} =$

$= \left\{ \begin{gathered}\ln \left( {1 - {x^2}} \right) = t, \hfill \\\frac{{ - 2x}}{{1 - {x^2}}}dx = dt \hfill \\ \end{gathered} \right\} = \frac{1}{2}\int {\frac{{dt}}{t}} = \frac{1}{2}\ln \left| t \right| + C = \frac{1}
{2}\ln \left| {\ln \left( {1 - {x^2}} \right)} \right| + C.$

3. Originally Posted by honestliar
This was our quiz last week which i failed(0/5), hah,, Can you show me how do I integrate this? What methods did my classmates used? Thanks.

$\int\frac{sin4xdx}{1+sin^22x}$
$\int {\frac{{\sin 4x}}{{1 + {{\sin }^2}2x}}dx} = \frac{1}{2}\int {\frac{{4\sin 2x\cos 2x}}{{1 + {{\sin }^2}2x}}dx} = \left\{ \begin{gathered}1 + {\sin ^2}2x = t, \hfill \\4\sin 2x\cos 2xdx = dt \hfill \\ \end{gathered} \right\} =$

$= \frac{1}{2}\int {\frac{{dt}}{t}} = \frac{1}{2}\ln \left| t \right| + C = \frac{1}{2}\ln \left( {1 + {{\sin }^2}2x} \right) + C.$

4. Originally Posted by honestliar
This was our quiz last week which i failed(0/5), hah,, Can you show me how do I integrate this? What methods did my classmates used? Thanks.

$\int\frac{\sqrt x dx}{1-\sqrt{x^3}}$
$\int {\frac{{\sqrt x }}{{1 - \sqrt {{x^3}} }}dx} = \int {\frac{{{x^{1/2}}}}{{1 - {x^{3/2}}}}dx} = - \frac{2}{3}\int {\frac{{ - \left( {3/2} \right){x^{1/2}}}}{{1 - {x^{3/2}}}}dx} =$

$= \left\{ \begin{gathered}1 - {x^{3/2}} = t, \hfill \\- \frac{3}
{2}{x^{1/2}}dx = dt \hfill \\ \end{gathered} \right\} = - \frac{2}
{3}\int {\frac{{dt}}{t}} = - \frac{2}{3}\ln \left| t \right| + C = - \frac{2}{3}\ln \left| {1 - {x^{3/2}}} \right| + C.$

5. Originally Posted by honestliar
This was our quiz last week which i failed(0/5), hah,, Can you show me how do I integrate this? What methods did my classmates used? Thanks.

$\int\frac{(2x-3) dx}{(x-2)^2}$
$\int {\frac{{2x - 3}}{{{{\left( {x - 2} \right)}^2}}}dx} = \int {\frac{{2\left( {x - 2} \right) + 1}}{{{{\left( {x - 2} \right)}^2}}}dx} = 2\int {\frac{{dx}}{{x - 2}}} + \int {\frac{{dx}}{{{{\left( {x - 2} \right)}^2}}}} =2$ $\ln \left| {x - 2} \right| - \frac{1}{{x - 2}} + C.$

6. For the last one, use the sub: $u = 1-x \ \Rightarrow \ -du = dx$

To get: $-\int \frac{\left[3-4(1-u)\right]\left(1-u\right) du}{u^{\frac{1}{3}}} \ = \ \cdots \ = \ \int \frac{4u^2 - 5u + 1}{u^{\frac{1}{3}}}du$

Divide through using exponent laws and it'll be a breeze.