Question

Integrate $\int \frac{x+e^x(\sin x+\cos x)+\sin x\cos x}{{(x^2+2e^x\sin x-{\cos }^{2}x)}^2}dx$

• A. $-\frac{1}{2(x^2+2e^x\sin x+{\cos }^{2}x)}+c$
• B. $-\frac{1}{2(x^2+2e^x\sin x-{\cos }^{2}x)}+c$
• C. $-\frac{1}{2(x^2+2e^x\cos x-{\sin }^{2}x)}+c$
• D. $-\frac{1}{2(x^2+2e^x\cos x+{\sin }^{2}x)}+c$

Answer 1

The correct option is B $-\frac{1}{2(x^2+2e^x\sin x-{\cos }^{2}x)}+c$

Let $I=\int \frac{x+e^x(\sin x+\cos x)+\sin x\cos x}{{(x^2+2e^x\sin x-{\cos }^{2}x)}^2}dx$

$=\frac{1}{2}\int \frac{2x+2e^x(\sin x+\cos x)+2\sin x\cos x}{{(x^2+2e^x\sin x-{\cos }^{2}x)}^2}dx$

Substitute $x^2+2e^x\sin x-{\cos }^{2}x=t$

$\Rightarrow (2x+2(e^x\cos x+e^x\sin x)+2\sin x\cos x)dx=dt$

$\therefore I=\frac{1}{2}\int \frac{dt}{t^2}=-\frac{1}{2}.\frac{1}{t}+c=-\frac{1}{2(x^2+2e^x\sin x-{\cos }^{2}x)}+c$