# Integration by substitution, what am I doing wrong

#### Tweety

Using $$\displaystyle u = \sqrt{2x+1}$$, show that

$$\displaystyle \int x(\sqrt{2x+1} dx = \frac{1}{15}(2x+1)^{\frac{3}{2}} (3x-1) + c$$

I have done this so many times and keep getting the wrong answer, heres my working.

$$\displaystyle u = \sqrt{2x+1}$$

$$\displaystyle x = \frac{1}{2}(u^{2} -1)$$

$$\displaystyle \frac{dx}{du} = u$$

$$\displaystyle dx = u du$$

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

$$\displaystyle \frac{1}{2} \int (u^{4} -u^{2}) du$$

$$\displaystyle \frac{1}{2}\int (\frac{u^{5}}{5} - \frac{u^{3}}{3})$$

$$\displaystyle \frac{1}{2} (\frac{3u^{5} -5u^{3}}{15} )$$

$$\displaystyle \frac{1}{2} \times \frac{1}{15} ( 3u^{5} - 5u^{3})$$

From here I am unable to get the desired expression, any help suggestions as to where I am going wrong would be greatly appreciated.

thanks

#### simplependulum

MHF Hall of Honor
Using $$\displaystyle u = \sqrt{2x+1}$$, show that

$$\displaystyle \int x(\sqrt{2x+1} dx = \frac{1}{15}(2x+1)^{\frac{3}{2}} (3x-1) + c$$

I have done this so many times and keep getting the wrong answer, heres my working.

$$\displaystyle u = \sqrt{2x+1}$$

$$\displaystyle x = \frac{1}{2}(u^{2} -1)$$

$$\displaystyle \frac{dx}{du} = u$$

$$\displaystyle dx = u du$$

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

$$\displaystyle \frac{1}{2} \int (u^{4} -u^{2}) du$$

$$\displaystyle \frac{u^{5}}{5} - \frac{u^{3}}{3}$$

$$\displaystyle \frac{1}{2} (\frac{3u^{5} -5u^{3}}{15} )$$

$$\displaystyle \frac{1}{2} \times \frac{1}{15} ( 3u^{5} - 5u^{3})$$

From here I am unable to get the desired expression, any help suggestions as to where I am going wrong would be greatly appreciated.

thanks
By factoring out $$\displaystyle u^3$$ in the last line we have

$$\displaystyle \frac{1}{30} (u^3)(3u^2-5)$$

Since $$\displaystyle u=\sqrt{2x+1}$$

$$\displaystyle = \frac{1}{30} \sqrt{2x+1}^3 ( 3 (2x+1) - 5)$$

$$\displaystyle = \frac{1}{30} \sqrt{2x+1}^3 ( 6x+ 3-5)$$

$$\displaystyle = \frac{1}{30} \sqrt{2x+1}^3 ( 6x-2)$$

$$\displaystyle =\frac{1}{30} \sqrt{2x+1}^3 [2( 3x-1 ) ]$$

$$\displaystyle = \frac{1}{15}\sqrt{2x+1}^3 ( 3x-1 )$$

Tweety

#### Prove It

MHF Helper
Using $$\displaystyle u = \sqrt{2x+1}$$, show that

$$\displaystyle \int x(\sqrt{2x+1} dx = \frac{1}{15}(2x+1)^{\frac{3}{2}} (3x-1) + c$$

I have done this so many times and keep getting the wrong answer, heres my working.

$$\displaystyle u = \sqrt{2x+1}$$

$$\displaystyle x = \frac{1}{2}(u^{2} -1)$$

$$\displaystyle \frac{dx}{du} = u$$

$$\displaystyle dx = u du$$

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

$$\displaystyle \frac{1}{2} \int (u^{4} -u^{2}) du$$

$$\displaystyle \frac{1}{2}\int (\frac{u^{5}}{5} - \frac{u^{3}}{3})$$

$$\displaystyle \frac{1}{2} (\frac{3u^{5} -5u^{3}}{15} )$$

$$\displaystyle \frac{1}{2} \times \frac{1}{15} ( 3u^{5} - 5u^{3})$$

From here I am unable to get the desired expression, any help suggestions as to where I am going wrong would be greatly appreciated.

thanks
$$\displaystyle \int{x\sqrt{2x + 1}\,dx}$$

Let $$\displaystyle u = 2x + 1$$, then $$\displaystyle x = \frac{u - 1}{2}$$ and $$\displaystyle \frac{du}{dx} = 2$$.

Go from here. Remember that $$\displaystyle \sqrt{u} = u^{\frac{1}{2}}$$.