Integration with exponential functions

**Bashyboy** · May 6th 2012, 02:37 PM

The problem is: $\int\frac{3^{2x}}{1+3^{2x}}\dx$

The answer I get is: $1/2\ln|1+3^{2x}|+C$ by using the formula $\int\frac{u'}{u}\du = \ln|u|+C$

Yet the answer is: $\frac{\ln(3^{2x}+1)}{2\ln3}+C$

I am not sure how how the natural logarithm in the denominator. Would it be possible for someone to help me?

**Plato** · May 6th 2012, 03:09 PM

Originally Posted by Bashyboy

The problem is: $\int\frac{3^{2x}}{1+3^{2x}}\dx$

The answer I get is: $1/2\ln|1+3^{2x}|+C$ by using the formula $\int\frac{u'}{u}\du = \ln|u|+C$

Yet the answer is: $\frac{\ln(3^{2x}+1)}{2\ln3}+C$

I am not sure how how the natural logarithm in the denominator. Would it be possible for someone to help me?

The derivative $D_x\left(3^{2x}\right)=2\left(3^{2x}\right)\ln(3)$ .

**HallsofIvy** · May 6th 2012, 03:38 PM

Originally Posted by Bashyboy

The problem is: $\int\frac{3^{2x}}{1+3^{2x}}\dx$

The answer I get is: $1/2\ln|1+3^{2x}|+C$ by using the formula $\int\frac{u'}{u}\du = \ln|u|+C$

I presume the "u" is in the denominator because you let $u= 1+ 3^{2x}$ . From that, $du= (2x)3^{2x} ln(2)dx$ .
Notice the "ln(2)" in the derivative. While it is true that the derivative of $e^x$ is $e^x$ , that is a special property of e. For any other base, say $a^x$ we can use the fact that ln(x) is the inverse of $e^x$ : $a^x= e^{ln(a^x)}= e^{x ln(a)}$ so that the derivative of $a^x$ is $e^{x ln(a)}$ times the derivative of x ln(a) which is ln(a). That is, the derivative of $a^x= e^{x ln(a)}$ is $ln(a)e^{xln(a)}= ln(a) a^x$ .

In other words, if $u= 3^{2x}$ , then $du= 2 ln(3) 3^{2x}dx$ so that $3^{2x}dx= \frac{du}{2 ln(3)}$

Yet the answer is: $\frac{\ln(3^{2x}+1)}{2\ln3}+C$

I am not sure how how the natural logarithm in the denominator. Would it be possible for someone to help me?

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