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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

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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 : $${\displaystyle\int}\dfrac{a^x}{x}\,\mathrm{d}x=\operatorname{E_i}\left(\ln\left(a\right)x\right)$$

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  • $\begingroup$ I shall look into this exponential integral..... But do u have any other efficient way of solving this out? $\endgroup$
    – Suraj S
    May 5 '17 at 18:56
  • $\begingroup$ Like using certain substitution and certain techniques? $\endgroup$
    – Suraj S
    May 5 '17 at 18:56
  • $\begingroup$ Thanks for the response...... $\endgroup$
    – Suraj S
    May 5 '17 at 18:57
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    $\begingroup$ @SurajS This seems to be the most efficient way of solving this. $\endgroup$ May 5 '17 at 19:00
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    $\begingroup$ 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 $\endgroup$
    – Bérénice
    May 5 '17 at 19:01

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