Find the value of x > 1 satisfying ∫x1 t ln t dt =1/4 is

The correct option is √e.

To find the value of $x>1$ that satisfies the given equation, we start by evaluating the definite integral on the left-hand side:
${\int }_1^{x}t\ln tdt=\frac{1}{4}$

We can evaluate this integral using the method of integration by parts. The formula for integration by parts is:
$\int udv=uv-\int vdu$

Let us choose:
$u=\ln t$
$dv=tdt$

Differentiating $u$ and integrating $dv$ gives:
$du=\frac{1}{t}dt$
$v=\frac{t^2}{2}$

Applying the integration by parts formula to the indefinite integral, we get:
$\int t\ln tdt=(\ln t)\left(\frac{t^2}{2}\right)-\int \left(\frac{t^2}{2}\right)\left(\frac{1}{t}dt\right)$
$=\frac{t^2\ln t}{2}-\frac{1}{2}\int tdt$
$=\frac{t^2\ln t}{2}-\frac{t^2}{4}$

Now, we apply the integration limits from 1 to $x$:
${\left[\frac{t^2\ln t}{2}-\frac{t^2}{4}\right]}_1^x=\frac{1}{4}$

Substitute the upper limit $x$ and the lower limit 1 into the expression:
$\left(\frac{x^2\ln x}{2}-\frac{x^2}{4}\right)-\left(\frac{1^2\ln 1}{2}-\frac{1^2}{4}\right)=\frac{1}{4}$

Since $\ln 1=0$, the lower limit evaluation simplifies to:
$-\left(0-\frac{1}{4}\right)=\frac{1}{4}$

Thus, our equation becomes:
$\frac{x^2\ln x}{2}-\frac{x^2}{4}+\frac{1}{4}=\frac{1}{4}$

Subtracting $\frac{1}{4}$ from both sides of the equation yields:
$\frac{x^2\ln x}{2}-\frac{x^2}{4}=0$

We can factor out $\frac{x^2}{2}$ from the expression:
$\frac{x^2}{2}\left(\ln x-\frac{1}{2}\right)=0$

Since we are given that $x>1$, we know that $x^2\ne 0$. Therefore, we can divide both sides by $\frac{x^2}{2}$:
$\ln x-\frac{1}{2}=0$
$\ln x=\frac{1}{2}$

To solve for $x$, we rewrite the logarithmic equation in exponential form:
$x=e^{1/2}$
$x=\sqrt{e}$

Since $e\approx 2.718$, we have $\sqrt{e}\approx 1.65$, which satisfies the condition $x>1$. Hence, the value of $x$ is indeed √e.