Question

If $I=\int (\frac{1}{x^{1/2}-x^{1/4}}+\frac{\log (1+x^{1/4})}{x^{1/2}+x^{1/4}})dx$, then I equals

• A. $2\sqrt{x}+4x^{1/4}+\log |x^{1/4}-1|+C$
• B. $2(x^{1/2}-6x^{1/4}+3)\log (x^{1/4}+1)+C-(x^{1/2}-10x^{1/4}+7)+2\left(\log \right(x^{1/4}+1){)}^2$
• C. $\sqrt{x}-x^{1/4}+6\log (x^{1/4}+1)-3x^{3/4}+C$
• D. none of these

Answer 1

The correct option is D none of these

$I=\int (\frac{1}{x^{1/2}-x^{1/4}}+\frac{\log (1+x^{1/4})}{x^{1/2}+x^{1/4}})dx$
Put $x=t^4$
$I=\int (\frac{4t^{3}dt}{t^2-t}+\frac{\log (1+t)}{t^2+1}4t^3)dt$
Let $I=I_1+I_2$
Where
$I_1=4\int \frac{t^2}{t-1}dt=4\int (t+1+\frac{1}{t-1})dt=2t^2+4t+\log |t-1|=2\sqrt{x}+4x^{3/4}+\log |x^{1/4}-1|$
And
$I_2=4\int \frac{\log (1+t)}{t+1}{t}^{2}dtt+1=u{I}_2=4\int \frac{\log u}{u}{(u-1)}^{2}du=4\int (u-2+\frac{1}{u})\log udu=(2u^2-8u)\log u-(u^2-8u)+2{\left(\log u\right)}^2=2(x^{1/2}-6x^{1/4}+3)\log (x^{1/4}+1)-(x^{1/2}-10x^{1/4}+7)+2{\left(\log (x^{1/4}+1)\right)}^2$