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# Evaluate the following integrals:

Question:

Evaluate the following integrals:

$\int \frac{x}{\sqrt{x^{4}+a^{4}}} d x$

Solution:

$\int \frac{\mathrm{x}}{\sqrt{\mathrm{x}^{4}+\mathrm{a}^{4}}} \mathrm{dx}=\int \frac{\mathrm{x}}{\sqrt{\left(\mathrm{x}^{2}\right)^{2}+\left(\mathrm{a}^{2}\right)^{2}}} \mathrm{dx}$

Let $x^{2}=t$, so $2 x d x=d t$

Or, $x d x=d t / 2$

Hence, $\int \frac{\mathrm{x}}{\sqrt{\left(\mathrm{x}^{2}\right)^{2}+\left(\mathrm{a}^{2}\right)^{2}}} \mathrm{dx}=\int \frac{1}{\sqrt{\mathrm{t}^{2}+\left(\mathrm{a}^{2}\right)^{2}}} \frac{\mathrm{dt}}{2}=\frac{1}{2} \int \frac{1}{\sqrt{\mathrm{t}^{2}+\left(\mathrm{a}^{2}\right)^{2}}} \mathrm{dt}$

Since, $\int \frac{1}{\sqrt{\left(x^{2}+a^{2}\right)}} d x=\log \left[x+\sqrt{\left(x^{2}+a^{2}\right)+c}\right.$

Hence, $\frac{1}{2} \int \frac{1}{\sqrt{t^{2}+\left(a^{2}\right)^{2}}} d t=\frac{1}{2} \log \left(t+\sqrt{t^{2}+\left(a^{2}\right)^{2}}+c\right.$

Put $t=x^{2}$

$=\frac{1}{2} \log \left(x^{2}+\sqrt{\left(x^{2}\right)^{2}+\left(a^{2}\right)^{2}}+c\right.$

$=\frac{1}{2} \log \left[x^{2}+\sqrt{x^{4}+a^{4}}\right]+c$