Evaluation of $\displaystyle \int\frac{1}{x^{\frac{1}{3}}+x^{\frac{1}{4}}}dx+\int\frac{\ln(1+x^{\frac{1}{6}})}{x^{\frac{1}{2}}+x^{\frac{1}{3}}}dx$

Question

Evaluation of $\displaystyle \int\frac{1}{x^{\frac{1}{3}}+x^{\frac{1}{4}}}dx+\int\frac{\ln(1+x^{\frac{1}{6}})}{x^{\frac{1}{2}}+x^{\frac{1}{3}}}dx$

$\bf{My\; Try::}$ Let $\displaystyle I = \int\frac{1}{x^{\frac{1}{3}}+x^{\frac{1}{4}}}dx\;,$ Now Put $x=t^{12}\;,$ Then $dx = 12t^{11}dt$

So we get $$I = 12\int\frac{t^{11}}{t^4+t^3}dt = 12\int \frac{t^8}{1+t}dt = 12\int\frac{(t^8-1)+1}{1+t}dt$$

So we get $$I = 12 \int (1+t+t^2+t^3+....+t^7)dt+12\ln |1+t|$$

and $\displaystyle J = \int\frac{\ln(1+x^{\frac{1}{6}})}{x^{\frac{1}{2}}+x^{\frac{1}{3}}}dx\;,$ Now put $x=u^6\;,$ We get $dx = 6u^5dt$

So we get $$ J = \int\frac{\ln(1+u)}{u^3+u^2}\cdot 6u^5dt = 6\int \frac{u^2\ln(1+u)}{1+u}du$$

Now How can I solve Integral $J\;,$ after that

Help required

Thanks

Answer 1

Considering $$A=\int \frac{u^2\ln(1+u)}{1+u}du$$ let us change variable $1+u=x$ which makes $$A=\int \frac{(x-1)^2 \log (x)}{x} dx=\int\left(x \log (x)+\frac{\log (x)}{x}-2 \log (x)\right)\,dx$$ Integrating the pieces by parts $$\int x\log(x)\,dx=\frac{1}{2} x^2 \log (x)-\frac{x^2}{4}$$ $$\int \frac{\log (x)}{x}dx=\frac{\log ^2(x)}{2}$$ $$\int \log(x)\,dx=x \log (x)-x$$

Answer 2

Let $v=\ln(1+u)$ so $u=e^v-1$ and $dv=\frac{du}{1+u}$.

So $$J=6\int (e^v-1)^2 v dv$$

Or

$$J=6\int (e^{2v}-2e^v+1)v dv$$

$$J=6\int ve^{2v}-2ve^v+v dv$$ Which shouldn't be hard. Break this into three integrals, use integration by parts for the first two and the third is easy.