Evaluate the integral:
$\int \frac{\sqrt{16+(\log x)^{2}}}{x} d x$
Key points to solve the problem:
- Such problems require the use of method of substitution along with method of integration by parts. By method of integration by parts if we have $\int f(x) g(x) d x=f(x) \int g(x) d x-\int f^{\prime}(x)\left(\int g(x) d x\right) d x$
- To solve the integrals of the form: $\int \sqrt{a x^{2}+b x+c} d x$ after applying substitution and integration by parts we have direct formulae as described below:
$\int \sqrt{a^{2}-x^{2}} d x=\frac{x}{2} \sqrt{a^{2}-x^{2}}+\frac{a^{2}}{2} \sin ^{-1}\left(\frac{x}{a}\right)+C$
$\int \sqrt{x^{2}-a^{2}} d x=\frac{x}{2} \sqrt{x^{2}-a^{2}}-\frac{a^{2}}{2} \log \left|x+\sqrt{x^{2}-a^{2}}\right|+C$
$\int \sqrt{x^{2}+a^{2}} d x=\frac{x}{2} \sqrt{x^{2}+a^{2}}+\frac{a^{2}}{2} \log \left|x+\sqrt{x^{2}+a^{2}}\right|+C$
Let, $I=\int \frac{1}{x} \sqrt{16+(\log x)^{2}} d x$
Let, $\log x=t$
Differentiating both sides:
$\Rightarrow \frac{1}{x} d x=d t$
Substituting $(\log x)$ with $t$, we have:
We have:
$I=\int \sqrt{t^{2}+16} d t=\int \sqrt{t^{2}+4^{2}} d t$
As I match with the form:
$\int \sqrt{x^{2}+a^{2}} d x=\frac{x}{2} \sqrt{x^{2}+a^{2}}+\frac{a^{2}}{2} \log \left|x+\sqrt{x^{2}+a^{2}}\right|+C$
$\therefore l=\left\{\frac{t}{2} \sqrt{t^{2}+16}+\frac{16}{2} \log \left|t+\sqrt{t^{2}+16}\right|\right\}+C$
Putting the value of $\mathrm{t}$ back:
$\Rightarrow I=\frac{\log x}{2} \sqrt{(\log x)^{2}+16}+8 \log \left|\log x+\sqrt{(\log x)^{2}+16}\right|+C$
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