# What substitution can be helpful here in this given integral problem

I am a beginner in integral calculus and I came across an integral problem which stated as follows. EDIT :

$$\int a^x {\left(\ln(x) + \ln(a) \ln \left(\frac{x} {e}\right)\right) } dx$$ I tried out substituting $$u = a^x$$ because I have terms like $\ln a, a^x$ which appears in $$du = a^x \ln(a) dx.$$

But then I worked out with using integration by parts and finally got stuck.

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

$$=\int a^x \ln(x) dx + \ln(a) \int a^x (\ln(x) - 1) dx$$ $$=\int \frac{1} {\ln(a)} \ln \left(\frac{\ln(u)} {\ln(a)} \right)du + \int \left(\ln \left(\frac{\ln(u)} {\ln(a)} \right) - 1 \right) du$$ Now I am stuck.

I couldn't figure out where I went wrong in the process. Can someone please help me out. If possible can anybody give me more efficient way of solving this problem. Thanks for the help

You could do an integration by part earlier considering the function $x\mapsto \ln\left(x\right)+\ln\left(a\right)\left(\ln\left(x\right)-1\right)$as the function to derivate and $x\mapsto a^x$ as the function to integrate thus you get : $$\int a^x {\left(\ln(x) + \ln(a) \ln \left(\frac{x} {e}\right)\right) } dx=\dfrac{a^x\left(\ln\left(x\right)+\ln\left(a\right)\left(\ln\left(x\right)-1\right)\right)}{\ln\left(a\right)}-{\displaystyle\int}\dfrac{\left(\frac{\ln\left(a\right)}{x}+\frac{1}{x}\right)a^x}{\ln\left(a\right)}\,\mathrm{d}x\\=\dfrac{a^x\left(\ln\left(x\right)+\ln\left(a\right)\left(\ln\left(x\right)-1\right)\right)}{\ln\left(a\right)}-\class{steps-node}{\cssId{steps-node-1}{\dfrac{\ln\left(a\right)+1}{\ln\left(a\right)}}}{\displaystyle\int}\dfrac{a^x}{x}\,\mathrm{d}x$$ But you can't get the last integral in term of elementary functions, but you can use this special function, the Exponential integral you get : $${\displaystyle\int}\dfrac{a^x}{x}\,\mathrm{d}x=\operatorname{E_i}\left(\ln\left(a\right)x\right)$$

• I shall look into this exponential integral..... But do u have any other efficient way of solving this out? May 5 '17 at 18:56
• Like using certain substitution and certain techniques? May 5 '17 at 18:56
• Thanks for the response...... May 5 '17 at 18:57
• @SurajS This seems to be the most efficient way of solving this. May 5 '17 at 19:00
• Well integration by part seems like the technic the most adapted fort this problem. And you can't solve the last integral without using this function. But I will see if there is an effective substitution May 5 '17 at 19:01