Find the following integral: int frac{x+3}{sq | vedclass

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

Question

(A) This integral is of the form $\int \frac{(p x+q) d x}{\sqrt{a x^{2}+b x+c}}$. Let us express $x+3 = A \frac{d}{d x}(5-4 x-x^{2}) + B = A(-4-2 x) +

Vedclass · Accepted Answer

$-\sqrt{5-4 x-x^{2}}+\sin ^{-1} \frac{x+2}{3}+ C$

Vedclass · Answer

(A) This integral is of the form $\int \frac{(p x+q) d x}{\sqrt{a x^{2}+b x+c}}$. Let us express $x+3 = A \frac{d}{d x}(5-4 x-x^{2}) + B = A(-4-2 x) + B$.Equating the coefficients of $x$ and the constant terms,we get $-2 A = 1$ and $-4 A + B = 3$,which gives $A = -\frac{1}{2}$ and $B = 1$.Therefore,$\int \frac{x+3}{\sqrt{5-4 x-x^{2}}} d x = -\frac{1}{2} \int \frac{(-4-2 x) d x}{\sqrt{5-4 x-x^{2}}} + \int \frac{d x}{\sqrt{5-4 x-x^{2}}} = -\frac{1}{2} I_{1} + I_{2} \dots (1)$.For $I_{1}$,let $5-4 x-x^{2} = t$,then $(-4-2 x) d x = d t$. Thus,$I_{1} = \int \frac{d t}{\sqrt{t}} = 2 \sqrt{t} + C_{1} = 2 \sqrt{5-4 x-x^{2}} + C_{1} \dots (2)$.For $I_{2}$,$I_{2} = \int \frac{d x}{\sqrt{9-(x+2)^{2}}} = \sin^{-1} \frac{x+2}{3} + C_{2} \dots (3)$.Substituting $(2)$ and $(3)$ in $(1)$,we get $\int \frac{x+3}{\sqrt{5-4 x-x^{2}}} d x = -\sqrt{5-4 x-x^{2}} + \sin^{-1} \frac{x+2}{3} + C$.

Find the following integral: $\int \frac{x+3}{\sqrt{5-4 x-x^{2}}} d x$

Similar Questions

$\int(\sqrt{\tan x}+\sqrt{\cot x}) d x=$

Evaluate: $\int \frac{2 \cos x+1}{(2+\cos x)^2} d x - \frac{\sin x}{2+\cos x}$

If $\int \frac{\log (1+x^4)}{x^3} d x=f(x) \log \left(\frac{1}{g(x)}\right)+\tan ^{-1}(h(x))+c$,then $h(x)\left[f(x)+f\left(\frac{1}{x}\right)\right]=$

$\int e^{3x} \sin(4x-5) dx = $ . . . . . . $+ C$

If $\int(x+2) \sqrt{x^2-x+2} \, dx = \frac{1}{3} f(x) + \frac{5}{8} g(x) + \frac{35}{16} h(x) + c$, then $f(-1) + g(-1) + h\left(\frac{1}{2}\right) = $

Vedclass Test Series

Exam Paper Generator

Online Exam Module