1. ## TRICKY u substitution - are my answers correct

QUESTION ONE $\int z^3\sqrt{z^2 + 1}dz$

$u = z^2 + 1$

$\frac{du}{dz} = 2z$

$\frac{z^2}{2}\int u^\frac{1}{2}\ du$

$\frac{z^2}{2}.\frac{2}{3}u^\frac{3}{2}$

$\frac{z^2}{3}\sqrt{(z^2+1)^3} + c$

QUESTION TWO

$\int y\sqrt{y+1}\ dy$

$u = y + 1$

$\frac{du}{dy} = 1$

$y\int u^\frac{1}{2}\ du$

$\frac{2y}{3}u^\frac{3}{2} + c$

$\frac{2y}{3}\sqrt{(y+1)^3} + c$

QUESTION THREE
$\int \frac{x}{(x-1)^3}\ dx$

$u = x -1$

$\frac{du}{dx}=1$

$x\int u^3\ du$

$x\frac{u^4}{4} + c$

$\frac{x(x-1)^4}{4}+c$

QUESTION FOUR $\int \frac{x}{\sqrt{2x-1}}\ dx$

$\frac{du}{dx} = 2$

$\frac{x}{2}\int u^\frac{-1}{2}\ du$

$\frac{x}{2}.2u^\frac{1}{2} + c$

$x\sqrt{2x -1} + c$

If incorrect, please enlighten me with regards to the correct way to do these. sorry, I'm really bad at Maths

Thank you.

2. Hi differentiate
Originally Posted by differentiate
QUESTION ONE [tex]

$\frac{z^2}{2}\int u^\frac{1}{2}\ du$

$\frac{z^2}{2}.\frac{2}{3}u^\frac{3}{2}$
This step is wrong. You have to state z^2 in term of u before integrating it.

QUESTION TWO

$y\int u^\frac{1}{2}\ du$

$\frac{2y}{3}u^\frac{3}{2} + c$
This step is wrong. You have to state y in term of u before integrating it.

QUESTION THREE
$x\int u^3\ du$

$x\frac{u^4}{4} + c$
This step is wrong. You have to state x in term of u before integrating it.

QUESTION FOUR $\int \frac{x}{\sqrt{2x-1}}\ dx$

$\frac{du}{dx} = 2$

$\frac{x}{2}\int u^\frac{-1}{2}\ du$

$\frac{x}{2}.2u^\frac{1}{2} + c$

$x\sqrt{2x -1} + c$

If incorrect, please enlighten me with regards to the correct way to do these. sorry, I'm really bad at Maths

Thank you.
What is your u ? ( I suppose it's 2x-1). And you have the same mistake as previous questions.

3. ohh thanks songoku! I get it now

4. alright, I got question 2 and 3 correct now. The problem is question 1 and question 2.

Question One $\int z^3\sqrt{z^2 + 1}\ dz$

$u = z^2 + 1$

$z^3 = (u-1)^\frac{3}{2}$

$\int u^\frac{1}{2}(u-1)^\frac{3}{2}\ du$

I'm stuck now. What should I do next?

QUESTION FOUR $\int \frac{x}{\sqrt{2x-1}}\ dx$

$u = 2x - 1$

$\frac{u + 1 }{2} = x$

$\int \frac{\frac{u+1}{2}}{u^\frac{1}{2}}\ du$

$\int \frac{u+1}{2u^\frac{1}{2}}$

$\int \frac{u^\frac{1}{2}}{2} + \frac{1}{2}u^\frac{-1}{2}\ du$

$\frac{1}{3}u^\frac{3}{2} + u^\frac{1}{2} + c$

$\frac{1}{3}\sqrt{(2x-1)^3} + \sqrt{2x-1} + c$

5. Originally Posted by differentiate
QUESTION ONE $\int z^3\sqrt{z^2 + 1}dz$

$u = z^2 + 1$

$\frac{du}{dz} = 2z$

$\frac{z^2}{2}\int u^\frac{1}{2}\ du$
It's good until this step. Now substitute z^2. Do the same for question 2

6. hmm.. I still couldn't get question 1

7. Is this the way you do it?

$\int z^3(z^2+1)\ dz$

$z^3 = (u-1)^\frac{3}{2}$

$\int (u-1)^\frac{3}{2}u^\frac{1}{2}$

8. Hi differentiate
Originally Posted by differentiate
QUESTION ONE $\int z^3\sqrt{z^2 + 1}dz$

$u = z^2 + 1$

$\frac{du}{dz} = 2z$

$\frac{z^2}{2}\int u^\frac{1}{2}\ du$

$\frac{z^2}{2}\int u^\frac{1}{2}\ du$

$=\frac{u-1}{2}\int u^\frac{1}{2}\ du$

$=\frac{1}{2}\int (u-1)u^\frac{1}{2}\ du$

9. Thanks skeeter. so continuing on... could someone check this for me?

$= \frac{1}{2}\int u^\frac{3}{2} - u^\frac{1}{2}\ du$

$\frac{1}{2}\left [ \frac{2}{5}u^\frac{5}{2} - \frac{2}{3}u^\frac{3}{2} \right ]$

$\frac{1}{5}u^\frac{5}{2} - \frac{1}{3}u^\frac{3}{2}$

$\frac{1}{5}(z^2+1)^\frac{5}{2} - \frac{1}{3}(z^2+1)^\frac{3}{2} + c$

Hence, evaluate $\int_{0}^{\sqrt{2}} z^3\sqrt{z^2 +1}$

$[\frac{1}{5}(z^2+1)^\frac{5}{2} - \frac{1}{3}(z^2+1)^\frac{3}{2} ]_0^\sqrt{2}$

$= \frac{1}{5}(2+1)^\frac{5}{2} - \frac{1}{3}(2+1)^\frac{3}{2} - \frac{1}{5} + \frac{1}{3}$

$= \frac{\sqrt{3^5}}{5} - \frac{\sqrt{3^3}}{3} + \frac{2}{15}$

$= \frac{9\sqrt{3}}{5} - \sqrt{3} + \frac{2}{15}$

10. Originally Posted by differentiate
QUESTION ONE $\int z^3\sqrt{z^2 + 1}dz$

$u = z^2 + 1$

$\frac{du}{dz} = 2z$

$\frac{z^2}{2}\int u^\frac{1}{2}\ du$

$\frac{z^2}{2}.\frac{2}{3}u^\frac{3}{2}$

$\frac{z^2}{3}\sqrt{(z^2+1)^3} + c$

QUESTION TWO

$\int y\sqrt{y+1}\ dy$

$u = y + 1$

$\frac{du}{dy} = 1$

$y\int u^\frac{1}{2}\ du$

$\frac{2y}{3}u^\frac{3}{2} + c$

$\frac{2y}{3}\sqrt{(y+1)^3} + c$

QUESTION THREE $\int \frac{x}{(x-1)^3}\ dx$

$u = x -1$

$\frac{du}{dx}=1$

$x\int u^3\ du$

$x\frac{u^4}{4} + c$

$\frac{x(x-1)^4}{4}+c$

QUESTION FOUR $\int \frac{x}{\sqrt{2x-1}}\ dx$

$\frac{du}{dx} = 2$

$\frac{x}{2}\int u^\frac{-1}{2}\ du$

$\frac{x}{2}.2u^\frac{1}{2} + c$

$x\sqrt{2x -1} + c$

If incorrect, please enlighten me with regards to the correct way to do these. sorry, I'm really bad at Maths

Thank you.
Q.2

$\int{ y\sqrt{y+1}\, dy}$

Let $u = y + 1$ so that $\frac{du}{dy} = 1$.

Also note that $y = u - 1$.

So the integral becomes

$\int{(u - 1)\sqrt{u}\,\frac{du}{dy}\,dy}$

$= \int{(u - 1)u^{\frac{1}{2}}\,du}$

$= \int{u^{\frac{3}{2}} - u^{\frac{1}{2}}\,du}$

$= \frac{2}{5}u^{\frac{5}{2}} - \frac{2}{3}u^{\frac{3}{2}} + C$

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

$= \frac{2}{5}\sqrt{(y + 1)^5} - \frac{2}{3}\sqrt{(y + 1)^3} + C$.

11. Originally Posted by differentiate
QUESTION ONE $\int z^3\sqrt{z^2 + 1}dz$

$u = z^2 + 1$

$\frac{du}{dz} = 2z$

$\frac{z^2}{2}\int u^\frac{1}{2}\ du$

$\frac{z^2}{2}.\frac{2}{3}u^\frac{3}{2}$

$\frac{z^2}{3}\sqrt{(z^2+1)^3} + c$

QUESTION TWO

$\int y\sqrt{y+1}\ dy$

$u = y + 1$

$\frac{du}{dy} = 1$

$y\int u^\frac{1}{2}\ du$

$\frac{2y}{3}u^\frac{3}{2} + c$

$\frac{2y}{3}\sqrt{(y+1)^3} + c$

QUESTION THREE $\int \frac{x}{(x-1)^3}\ dx$

$u = x -1$

$\frac{du}{dx}=1$

$x\int u^3\ du$

$x\frac{u^4}{4} + c$

$\frac{x(x-1)^4}{4}+c$

QUESTION FOUR $\int \frac{x}{\sqrt{2x-1}}\ dx$

$\frac{du}{dx} = 2$

$\frac{x}{2}\int u^\frac{-1}{2}\ du$

$\frac{x}{2}.2u^\frac{1}{2} + c$

$x\sqrt{2x -1} + c$

If incorrect, please enlighten me with regards to the correct way to do these. sorry, I'm really bad at Maths

Thank you.
Q.3

$\int{\frac{x}{(x - 1)^3}\,dx}$.

Let $u = x - 1$ so that $\frac{du}{dx} = 1$.

Also note that $x = u + 1$.

Therefore, the integral becomes

$\int{\frac{u + 1}{u^3}\,\frac{du}{dx}\,dx}$

$= \int{(u + 1)u^{-3}\,du}$

$= \int{u^{-2} + u^{-3}\,du}$

$= -u^{-1} - \frac{1}{2}u^{-2} + C$

$= -\frac{1}{x - 1} - \frac{1}{2(x - 1)^2} + C$.

12. Originally Posted by differentiate
QUESTION ONE $\int z^3\sqrt{z^2 + 1}dz$

$u = z^2 + 1$

$\frac{du}{dz} = 2z$

$\frac{z^2}{2}\int u^\frac{1}{2}\ du$

$\frac{z^2}{2}.\frac{2}{3}u^\frac{3}{2}$

$\frac{z^2}{3}\sqrt{(z^2+1)^3} + c$

QUESTION TWO

$\int y\sqrt{y+1}\ dy$

$u = y + 1$

$\frac{du}{dy} = 1$

$y\int u^\frac{1}{2}\ du$

$\frac{2y}{3}u^\frac{3}{2} + c$

$\frac{2y}{3}\sqrt{(y+1)^3} + c$

QUESTION THREE $\int \frac{x}{(x-1)^3}\ dx$

$u = x -1$

$\frac{du}{dx}=1$

$x\int u^3\ du$

$x\frac{u^4}{4} + c$

$\frac{x(x-1)^4}{4}+c$

QUESTION FOUR $\int \frac{x}{\sqrt{2x-1}}\ dx$

$\frac{du}{dx} = 2$

$\frac{x}{2}\int u^\frac{-1}{2}\ du$

$\frac{x}{2}.2u^\frac{1}{2} + c$

$x\sqrt{2x -1} + c$

If incorrect, please enlighten me with regards to the correct way to do these. sorry, I'm really bad at Maths

Thank you.
Q.4

$\int{\frac{x}{\sqrt{2x - 1}}\,dx}$.

Let $u = 2x - 1$ so that $\frac{du}{dx} = 2$.

Also note that $x = \frac{1}{2}(u + 1)$.

So the integral becomes

$\frac{1}{2}\int{\frac{2x}{\sqrt{2x - 1}}\,dx}$

$= \frac{1}{2}\int{\frac{u + 1}{2\sqrt{u}}\,\frac{du}{dx}\,dx}$

$= \frac{1}{4}\int{(u + 1)u^{-\frac{1}{2}}\,du}$

$= \frac{1}{4}\int{u^{\frac{1}{2}} + u^{-\frac{1}{2}}\,du}$

$= \frac{1}{4}\left[\frac{2}{3}u^{\frac{3}{2}} + 2u^{\frac{1}{2}} + c\right]$

$= \frac{1}{6}(2x - 1)^{\frac{3}{2}} + \frac{1}{2}(2x - 1)^{\frac{1}{2}} + C$

$= \frac{1}{6}\sqrt{(2x - 1)^3} + \frac{1}{2}\sqrt{2x - 1} + C$.