int frac{dx}{(x+1) sqrt{4x+3}} is equal to | vedclass

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

Question

(C) Let $I = \int \frac{dx}{(x+1) \sqrt{4x+3}}$.Substitute $4x+3 = t^2$,which implies $4dx = 2tdt$ or $dx = \frac{1}{2} t dt$.Also,$x = \frac{t^2-3}{4

Vedclass · Accepted Answer

$2 	an^{-1} \sqrt{4x+3} + c$

Vedclass · Answer

(C) Let $I = \int \frac{dx}{(x+1) \sqrt{4x+3}}$.Substitute $4x+3 = t^2$,which implies $4dx = 2tdt$ or $dx = \frac{1}{2} t dt$.Also,$x = \frac{t^2-3}{4}$,so $x+1 = \frac{t^2-3}{4} + 1 = \frac{t^2+1}{4}$.Substituting these into the integral:$I = \int \frac{\frac{1}{2} t dt}{(\frac{t^2+1}{4}) t} = \int \frac{\frac{1}{2} dt}{\frac{t^2+1}{4}} = 2 \int \frac{dt}{t^2+1}$.Integrating,we get $I = 2 	an^{-1}(t) + c$.Substituting back $t = \sqrt{4x+3}$,we get $I = 2 	an^{-1} \sqrt{4x+3} + c$.

$\int \frac{dx}{(x+1) \sqrt{4x+3}}$ is equal to

Similar Questions

The absolute value of the tangent of the difference of the angles made by the lines $4x^2 - 24xy + 11y^2 = 0$ with the $X$-axis is

The point on the line $4x - y - 2 = 0$ which is equidistant from the points $(-5, 6)$ and $(3, 2)$ is

In biennial plants,normally the number of carpels is:

Two sources of equal emf $E$ are connected in series to an external resistance $R$. The internal resistances of the two sources are $R_1$ and $R_2$ $(R_2 > R_1)$. If the potential difference across the source having internal resistance $R_2$ is zero,then the value of $R$ is:

The output characteristics of an $n-p-n$ transistor represent,($I_C=$ collector current,$V_{C E}=$ potential difference between collector and emitter,$I_B=$ base current,$V_{B B}=$ voltage given to base,$V_{B E}=$ the potential difference between base and emitter)

Vedclass Test Series

Exam Paper Generator

Online Exam Module