Question
Evaluate the following integrals:
$\int\text{x}\sqrt{\text{x}^4+1}\text{dx}$
$\int\text{x}\sqrt{\text{x}^4+1}\text{dx}$
✓
Answer
$\text{I}=\int\text{x}\sqrt{\text{x}^4+1}\text{dx}$
$=\int\text{x}\sqrt{(\text{x}^2)^2+1}\text{dx}$
Putting $\text{x}^2=\text{t}$
$\Rightarrow2\text{x dx}=\text{dt}$
$\Rightarrow\text{x dx}=\frac{\text{dt}}{2}$
$\therefore\ \text{I}=\frac{1}{2}\int\sqrt{\text{t}^2+1}\text{dt}$
$=\frac{1}{2}\int\sqrt{\text{t}^2+1}\text{dt}$
$=\frac{1}{2}\Big[\frac{\text{t}}{2}\sqrt{\text{t}^2+1}+\frac{1^2}{2}\log\big|\text{t}+\sqrt{\text{t}^2+1}\big|\Big]+\text{C}$
$=\frac{1}{2}\Big[\frac{\text{x}^2}{2}\sqrt{\text{x}^4+1}+\frac{1}{2}\log\big|\text{x}^2+\sqrt{\text{x}^4+1}\big|\Big]+\text{C}$
$=\frac{\text{x}^2}{4}\sqrt{\text{x}^4+1}+\frac{1}{4}\log\big|\text{x}^2+\sqrt{\text{x}^4+1}+\text{C}$
$=\int\text{x}\sqrt{(\text{x}^2)^2+1}\text{dx}$
Putting $\text{x}^2=\text{t}$
$\Rightarrow2\text{x dx}=\text{dt}$
$\Rightarrow\text{x dx}=\frac{\text{dt}}{2}$
$\therefore\ \text{I}=\frac{1}{2}\int\sqrt{\text{t}^2+1}\text{dt}$
$=\frac{1}{2}\int\sqrt{\text{t}^2+1}\text{dt}$
$=\frac{1}{2}\Big[\frac{\text{t}}{2}\sqrt{\text{t}^2+1}+\frac{1^2}{2}\log\big|\text{t}+\sqrt{\text{t}^2+1}\big|\Big]+\text{C}$
$=\frac{1}{2}\Big[\frac{\text{x}^2}{2}\sqrt{\text{x}^4+1}+\frac{1}{2}\log\big|\text{x}^2+\sqrt{\text{x}^4+1}\big|\Big]+\text{C}$
$=\frac{\text{x}^2}{4}\sqrt{\text{x}^4+1}+\frac{1}{4}\log\big|\text{x}^2+\sqrt{\text{x}^4+1}+\text{C}$
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