What is the equation of the tangent drawn to the function $f(x)$ at the point $ x = 1 $

Question

Here is question and my attempts: I can not get the correct answer.

Let $$f(x)=\int_x^{x^2}\frac{2t^2+1}{t^3+2}dt$$ What is the equation of the tangent drawn to the function $f(x)$ at the point $ x = 1 $?

A) $y=-x$

B) $y=x$

C) $y=2x$

D) $y=\frac x2$

E) $y=\frac{3x}2$

My attempt: \begin{align} f(x)&=\int_x^{x^2}\frac{2t^2+1}{t^3+2}dt\\ y&=f(1)+f'(1)(x-1)\\ f(1)&=\int_1^{1}\frac{2t^2+1}{t^3+2}dt=0\\ f'(x)&=\frac{2x^4+1}{x^6+2}\cdot2x-\frac{2x^2+1}{x^3+2}\cdot1\\ f'(1)&=\frac33\cdot2-\frac33\cdot1=1\\ y&=x-1 \end{align}

Answer 1

Your solution is totally fine. Except for a lack of words I have nothing to say about it. Your textbook is incorrect, as the correct answer is not part of the options.

Answer 2

As you notice, none of the given answers satisfy $f(1)=0$ which is required.

I have checked your work and your solution is correct.

Answer 3

You are correct. The proposed answers are all incorrect because $f(1)=0$ and any line $y=mx$ with $ m\not=0$ gives $y(1)=m\not=0$.

Note that the tangent to the graph of $f$ at $x_0=0$ is $$y=f(0)+f'(0)x=-\frac{x}{2}.$$ So even for such point all answers are wrong.