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# Evaluate : $\int \large\frac{2x+5}{\sqrt{7-6x-x^2}}$$dx Can you answer this question? ## 1 Answer 0 votes Toolbox: • \int\large\frac{(px+q)}{\sqrt{ax^2+bx+c}}$$dx$,where p,q,a,b,c are constants,we are to find real numbers such that $px+q=A\frac{d}{dx}(ax^2+bx+c)+B.$
• Therefore $px+q=A(2ax+b)+B.$
• To determine A and B we equate the coefficients from both sided,the coefficients of x and the constant terms.A and B are thus obtained and hence the integral is reduced to one of the known forms
• $\int \large\frac{dx}{\sqrt{a^2-x^2}}$$=-\sin^{-1}(\large\frac{x}{a}) Step 1: I=\int\large\frac{2x+5}{\sqrt{7-6x-x^2}}$$dx$
Let us express
$2x+5=A\large\frac{d}{dx}$$(7-6x-x^2)+B 2x+5=A(-6-2x)+B Now equating the coefficients of x and the constant terms from both sides we get, -2A=2 A=-1 -6A+B=5 (i.e) -6(-1)+B=5 6+B=5 B=-1 Hence A=-1 and B=-1 \therefore \int\large\frac{2x+5}{\sqrt{7-6x-x^2}}=\int\large\frac{-1(-6-2x)dx}{\sqrt{7-6x-x^2}}+\int\large\frac{-dx}{\sqrt{7-6x-x^2}} \qquad\qquad\qquad=I_1-I_2 Step 2: In I_1 put 7-6x-x^2=t \Rightarrow (-6-2x)dx=dt \therefore I_1=-\int\large\frac{dt}{\sqrt t} \qquad=-2\sqrt t+C_1 \qquad=-2\sqrt {7-6x-x^2}+C_1 I_2=-\int\large\frac{dx}{\sqrt{7-6x-x^2}} \quad=-\int\large\frac{dx}{\sqrt{16-(x+3)^2}} \int \large\frac{dx}{\sqrt{a^2-x^2}}$$=-\sin^{-1}(\large\frac{x}{a})$