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If $\int \frac{x+1}{\sqrt{2x-1}}dx=f(x)\sqrt{2x-1}+C$ , where C is a constant of integration, then $f(x)$ is equal to :

Option: 1 $\frac{2}{3}(x+2)$
Option: 2 $\frac{1}{3}(x+1)$
Option: 3 $\frac{1}{3}(x+4)$
Option: 4 $\frac{2}{3}(x-4)$

Answers (1)

best_answer

Indefinite integrals for Algebraic functions -

$\frac{\mathrm{d}}{\mathrm{d} x} \frac{\left ( x^{n+1} \right )}{n+1}=x^{n}$ so $\int x^{n}dx=\frac{x^{n+1}}{n+1}$

- wherein

Where $n\neq-1$

Integration by substitution -

The functions when on substitution of the variable of integration to some quantity gives any one of standard formulas.

- wherein

Since $\int f(x)dx=\int f(t)dt=\int f(\theta )d\theta$ all variables must be converted into single variable , $\left ( t\, or\ \theta \right )$

Let t = $\sqrt{2x-1}$

$=>t^{2}={2x-1}$

$=>2t\: dt=2\: dx$

$\int \frac{x+1}{\sqrt{2x-1}}dx=\int \frac{t^{2}+3}{2}dt$

$=\frac{1}{2}\left [ \frac{t^{3}}{3}+3t \right ]$

$=\frac{t}{6}(t^{2}+9)+C$

t = $\sqrt{2x-1}$

$=\frac{\sqrt{2x-1}}{6}(2x+8)+C$

$=\frac{\sqrt{2x-1}}{3}(x+4)+C$

$=>f (x)=\frac{x+4}{3}$

Posted by

Ritika Jonwal

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