Câu 1:
\(A=\int \frac{2\sin x+\cos x}{3\sin x+2\cos x}dx\)
\(A=\int \frac{\frac{8}{13}(3\sin x+2\cos x)-\frac{1}{13}(3\cos x-2\sin x)}{3\sin x+2\cos x}dx\)
\(A=\frac{8}{13}\int dx-\frac{1}{13}\int \frac{(3\cos x-2\sin x)dx}{3\sin x+2\cos x}\)
\(A=\frac{8}{13}x-\frac{1}{13}\int \frac{d(3\sin x+2\cos x)}{3\sin x+2\cos x}\)
\(A=\frac{8}{13}x-\frac{1}{13}\ln |3\sin x+2\cos x|+c\)
Câu 2:
Ta có: \(I=\int \frac{x^3}{x^4+3x^2+2}dx=\int \frac{x^3}{(x^2+1)(x^2+2)}dx\)
\(=\int x^3\left(\frac{1}{x^2+1}-\frac{1}{x^2+2}\right)dx=\int \frac{x^3dx}{x^2+1}-\int \frac{x^3}{x^2+2}dx\)
\(=\frac{1}{2}\int \frac{x^2d(x^2+1)}{x^2+1}-\frac{1}{2}\int \frac{x^2d(x^2+2)}{x^2+2}\)
\(=\frac{1}{2}\int \left(1-\frac{1}{x^2+1}\right)d(x^2+1)-\frac{1}{2}\int \left(1-\frac{2}{x^2+2}\right)d(x^2+2)\)
\(=\frac{1}{2}\int d(x^2+1)-\frac{1}{2}\int \frac{d(x^2+1)}{x^2+1}-\frac{1}{2}\int d(x^2+2)+\int \frac{d(x^2+2)}{x^2+2}\)
\(=\frac{x^2+1}{2}-\frac{1}{2}\ln |x^2+1|-\frac{x^2+2}{2}+\ln |x^2+2|+c\)
\(=\ln |x^2+2|-\frac{1}{2}\ln |x^2+1|+c\)
