# Thread: Need some help with this integral

1. ## Need some help with this integral

I'm having trouble evaluating thhis integral

S = integral sign

S 4 ( 5^4x+9) dx

I tried using X^n = (X^n+1 / n+1) + C by taking the constant 4 outside and to the left but the result after applying the above rule made no sense.

I then tried using substitution where 4x+9 = alpha. So the derivative of alpha is just 4 but cannot reconstruct what this will give. Here's what I ended up with:

alpha = @= 4x+9
derivative @ = d@ = 4
d@ / 4 = dx

thus,

S 4 (5^4x+9) dx= (5^4x+9) d@/4 but now I don't know what to do.

2. $\int 4(5^{4x}+9) dx = 4 [\int 5^{4x} dx + \int 9 dx]$

$\int 5^{4x} dx$ can be integrated this way:

let $u = 5^4x$

so, $ln(u) = ln(5)^{4x} \implies ln(u) = 4x \times ln(5) \implies x = \frac{ln(u)}{4ln(5)}$

$\therefore dx = \frac{1}{4 ln(5)} \times \frac{1}{u} \mbox{du}$

substitute u and dx to get:

$\int 5^{4x} dx = \int u \frac{1}{4 ln(5)} \times \frac{1}{u} \mbox{du}$

and integrate to complete ....

3. The exponent of 5 is 4x+9.

Could I still use the ln(u) method for this?

4. Yes.
You can do the same as I have done above.

Start with $u = 5^{4x+9}$ and follow what I did previously(take log and differentiate).....

PS: Please make your notations clear. Use LaTEX if possible!

5. Your "power rule", that the anti-derivative of $x^n$ is $\frac{1}{n+1}x^{n+1}$ requires that x be the base not the exponent.

The derivative of $a^x$ is $ln(a) a^x$ and the anti-derivative is $\frac{a^x}{ln(a)}$ (which you can derive by taking logarithms as harish21 did).

To integrate $\int 5^{4x+9} dx$, let u= 4x+ 9 so that du= 4dx, dx= du/4. The integral becomes $\frac{1}{4}\int 5^u du= \frac{5^u}{4ln(5)}+ C= \frac{5^{4x+9}}{4ln(5)}+ C$.

6. An alternative method is to write

$\displaystyle{5^{4x+9} = e^{\ln{(5^{4x+9})}}}$

$\displaystyle{=e^{(4x+9)\ln{5}}}$

$\displaystyle{= e^{(4\ln{5})x + 9\ln{5}}}$.

So $\displaystyle{\int{4\cdot 5^{4x+9}\,dx} = \int{4e^{(4\ln{5})x + 9\ln{5}}\,dx}}$

$\displaystyle{= \frac{1}{\ln{5}}\int{(4\ln{5})e^{(4\ln{5})x + 9\ln{5}}\,dx}}$.

Now make the substitution $\displaystyle{u = (4\ln{5})x + 9\ln{5}}$ so that $\displaystyle{\frac{du}{dx} = 4\ln{5}}$ and the integral becomes

$\displaystyle{\frac{1}{\ln{5}}\int{(4\ln{5})e^{(4\ ln{5})x + 9\ln{5}}\,dx} = \frac{1}{\ln{5}}\int{e^u\,\frac{du}{dx}\,dx}}$

$\displaystyle{=\frac{1}{\ln{5}}\int{e^u\,du}}$

$\displaystyle{= \frac{1}{\ln{5}}\,e^u + C}$

$\displaystyle{=\frac{1}{\ln{5}}\,e^{(4\ln{5})x + 9\ln{5}} + C}$

$\displaystyle{ = \frac{1}{\ln{5}}\,e^{(4x + 9)\ln{5}} + C}$

$\displaystyle{= \frac{1}{\ln{5}}\,e^{\ln{(5^{4x + 9})}} + C}$

$\displaystyle{= \frac{5^{4x + 9}}{\ln{5}}} + C$.

7. Yep Thanks a lot guys I had the rules mixed up. Solved it. =) Pretty easy too :P