1. Write the sum clearly

Given

$$I=\sum_{k=1}^{98}\int_k^{k+1}\frac{k+1}{x(x+1)}\,dx$$

We evaluate the general term

$$I_k=\int_k^{k+1}\frac{k+1}{x(x+1)}\,dx.$$

2. Use partial fractions

Recall

$$\frac{1}{x(x+1)}=\frac{1}{x}-\frac{1}{x+1}.$$

So

$$I_k=(k+1)\int_k^{k+1}\left(\frac{1}{x}-\frac{1}{x+1}\right)dx.$$

3. Integrate

Thus

$$I_k=(k+1)\left[\ln x-\ln(x+1)\right]_k^{k+1}.$$

Now evaluate at the limits:

$$I_k=(k+1)\left(\ln\frac{k+1}{k+2}-\ln\frac{k}{k+1}\right)$$ $$=(k+1)\ln\left(\frac{(k+1)^2}{k(k+2)}\right).$$

Since

$$(k+1)^2=k(k+2)+1,$$

this becomes

$$I_k=(k+1)\ln\left(1+\frac{1}{k(k+2)}\right).$$

That form is not the easiest for summation, so we use another approach.

4. Compare the integrand with simpler functions

For $x\in[k,k+1]$, we have

$$x\le k+1 \implies \frac{k+1}{x}\ge 1,$$

and hence

$$\frac{k+1}{x(x+1)}\ge \frac{1}{x+1}.$$

Also,

$$x\ge k \implies \frac{k+1}{x}\le \frac{k+1}{k},$$

but a better comparison is:

$$x\ge k \implies \frac{k+1}{x}\le \frac{k+1}{k},$$

not immediately useful. Instead observe

$$x\le k+1 \implies x+1\le k+2,$$

so comparisons are easier directly:

Because $x\in[k,k+1]$,

$$k\le x\le k+1.$$

Thus

$$\frac{k+1}{x}\ge 1$$

which gives

$$\frac{k+1}{x(x+1)}\ge \frac{1}{x+1}.$$

Integrating from $k$ to $k+1$,

$$I_k\ge \int_k^{k+1}\frac{1}{x+1}\,dx=\ln\frac{k+2}{k+1}.$$

Also, since $x\ge k$,

$$\frac{k+1}{x}\le \frac{k+1}{k},$$

so

$$\frac{k+1}{x(x+1)}\le \frac{k+1}{k}\cdot \frac{1}{x+1},$$

which is not ideal. A sharper simple upper bound is obtained from

$$x+1\ge k+1 \implies \frac{1}{x+1}\le \frac{1}{k+1},$$

so

$$\frac{k+1}{x(x+1)}\le \frac{1}{x}.$$

Indeed,

$$\frac{k+1}{x(x+1)}\le \frac{1}{x} \iff k+1\le x+1 \iff x\ge k,$$

which is true.

Hence

$$I_k\le \int_k^{k+1}\frac{1}{x}\,dx=\ln\frac{k+1}{k}.$$

Therefore,

$$\ln\frac{k+2}{k+1}\le I_k\le \ln\frac{k+1}{k}.$$

5. Sum over $k=1$ to $98$

Summing the lower bounds:

$$I\ge \sum_{k=1}^{98}\ln\frac{k+2}{k+1} =\ln\prod_{k=1}^{98}\frac{k+2}{k+1}.$$

This telescopes:

$$\prod_{k=1}^{98}\frac{k+2}{k+1}=\frac{3}{2}\cdot\frac{4}{3}\cdots\frac{100}{99}=\frac{100}{2}=50.$$

So

$$I\ge \ln 50.$$

Summing the upper bounds:

$$I\le \sum_{k=1}^{98}\ln\frac{k+1}{k} =\ln\prod_{k=1}^{98}\frac{k+1}{k} =\ln 99.$$

In fact the inequality is strict, because on each interval $[k,k+1]$,

$$\frac{k+1}{x(x+1)}<\frac{1}{x}$$

for all $x>k$ (except only at the single point $x=k$), hence

$$I_k<\int_k^{k+1}\frac{1}{x}\,dx.$$

Therefore

$$I<\ln 99.$$

So option B is correct and option A is false.

6. Check against $\dfrac{49}{50}$

From the lower bound,

$$I\ge \ln 50.$$

Now

$$\ln 50>1>\frac{49}{50}.$$

Hence certainly

$$I>\frac{49}{50}.$$

So option D is correct and option C is false.

7. Final conclusion

Correct options are:

$$\boxed{\text{B, D}}$$