Let's do some derivative #2

$\Rightarrow \displaystyle\int_0^{x^2} ( 1 + t)\color{#3D99F6}{f}(t) \text{dt} = 6x^4$

Applying Second fundamental theorem of calculus ,

$\Rightarrow ( 1 + x^2)\color{#3D99F6}{f}(x^2)(2x) = 24x^3$

$\Rightarrow \color{#3D99F6}{f}(\color{#D61F06}{x^2})= \dfrac{12\color{#D61F06}{x^2}}{1 + \color{#D61F06}{x^2}}$

Substituting $\color{#D61F06}{x^2} = 1$

$\Rightarrow \color{#3D99F6}{f(1)} = 6$

Assuming the fundamental theorem of Calculus and, that two differentiable functions in (a, b) such that they have the same derivative differ by a constant. With this in mind, the functions of both members of the equality vanish at x = 0. Then, It is sufficient to match the derivatives ( $1 + x^2$ )f( $x^2$ )2x = 24 $x^3$ . (Now, $x\ge 0$ ) $\Rightarrow$ f(x)= 12x / (1 + x) $\Rightarrow$ f(1) = 6

It is very simple to show that $\int_{0}^{x^2} 12t dt =6x^4$ comparing it with the question , we can find f(t), $f(t) = \frac{12t}{1+t}$ Hence $f(1)=6$

So we're given the following equation.

$\int _{ 0 }^{ { x }^{ 2 } }{ (1+t)f(t)\quad dt\quad=\quad 6{ x }^{ 4 } }$

Since $t\quad=\quad { x }^{ 2 }$ we can rewrite this in terms of t as follows.

$\int _{ 0 }^{ t }{ (1+t)f(t)\quad dt\quad=\quad 6{ t }^{ 2 } }$

Then we can take the derivative of f with respect to t on both sides

$\frac { df }{ dt } \int _{ 0 }^{ t }{ (1+t)f(t)\quad dt\quad =\quad \frac { df }{ dt } 6{ t }^{ 2 } }$

$\Longrightarrow$ $(1+t)f(t)\quad =\quad 12t$

$\Longrightarrow$ $f(t)=\frac { 12t }{ 1+t }$

$\Longrightarrow$ $f(1)=\frac { 12 }{ 2 } =6$