JEE Main 2025 — 28 January, Evening Shift — Question 11

let $\mathrm{x}=\mathrm{t}^{4}$

$$\mathrm{dx}=4 \mathrm{t}^{3} \mathrm{dt}$$

then $\int \frac{1}{x^{\frac{1}{4}}\left(1+x^{\frac{1}{4}}\right)} d x \Rightarrow \int \frac{4 t^{3} d t}{t(1+t)}$

$\Rightarrow \int \frac{4 \mathrm{t}^{2}}{1+\mathrm{t}} \mathrm{dt} \Rightarrow 4 \int \frac{\left(\mathrm{t}^{2}-1\right)+1}{1+\mathrm{t}} \mathrm{dt}$

$\Rightarrow 4 \int(\mathrm{t}-1)+\frac{1}{\mathrm{t}+1} \mathrm{dt}$

$\Rightarrow 4\left\{\frac{(\mathrm{t}-1)^{2}}{2}+\ln (\mathrm{t}+1)\right\}+\mathrm{c}$

hence $f(x)=2\left(x^{\frac{1}{4}}-1\right)^{2}+4 \ell n\left(1+x^{\frac{1}{4}}\right)+c$

$f(0)=-6 \Rightarrow 2+C=-6 \rightarrow C=-8$

now $f(1)=4 \ell$ n $2-8$

$$=4(\ln 2-2)$$