If fleft(frac{t+1}{2 t+1}right)=t+1, then int | vedclass

Name: Exam Paper Generator | JEE NEET Paper Generator | Vedclass
Brand: Vedclass
Rating: 5 (35 reviews)

Question

(D) We have, $f\left(\frac{t+1}{2 t+1}\right)=t+1$. Let $\frac{t+1}{2 t+1}=x$. Then $t+1=x(2t+1) \Rightarrow t+1=2tx+x$. $t(1-2x)=x-1 \Rightarrow t=\f

Vedclass · Accepted Answer

$\frac{x}{2}+\frac{1}{4} \log (2 x-1)+c$

Vedclass · Answer

(D) We have, $f\left(\frac{t+1}{2 t+1}\right)=t+1$. Let $\frac{t+1}{2 t+1}=x$. Then $t+1=x(2t+1) \Rightarrow t+1=2tx+x$. $t(1-2x)=x-1 \Rightarrow t=\frac{x-1}{1-2x}=\frac{1-x}{2x-1}$. Substituting $t$ into the expression for $f(x)$: $f(x) = \frac{1-x}{2x-1} + 1 = \frac{1-x+2x-1}{2x-1} = \frac{x}{2x-1}$. Now, we calculate the integral: $\int f(x) dx = \int \frac{x}{2x-1} dx = \frac{1}{2} \int \frac{2x}{2x-1} dx$. $= \frac{1}{2} \int \frac{2x-1+1}{2x-1} dx = \frac{1}{2} \int \left(1 + \frac{1}{2x-1}\right) dx$. $= \frac{1}{2} \left[ x + \frac{1}{2} \log |2x-1| \right] + c$. $= \frac{x}{2} + \frac{1}{4} \log |2x-1| + c$.

If $f\left(\frac{t+1}{2 t+1}\right)=t+1$, then $\int f(x) d x=$

Similar Questions

Find: $\int \cos^{2} x \, dx$

The value of $\int \frac{1}{1+\cos 8x} dx$ is

The velocity of a particle at time $t$ is given by the relation $v = 6t - \frac{t^2}{6}$. The distance traveled in $3 \ s$ is,if $s = 0$ at $t = 0$.

Consider the following statements $(A)$ and $(B)$:
$(A) \int_a^b \frac{d}{d x}(f(x)) d x = \frac{d}{d x} \int_a^b f(x) d x$
$(B) \frac{d}{d x} \left( \int f(x) d x \right) = f(x) + C$
Which one of the following is true?

$\int \frac{dx}{\sin x+\sqrt{3} \cos x}$ equals, where $c$ is an arbitrary constant.

Vedclass Test Series

Exam Paper Generator

Online Exam Module