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(1+x)sin(4x)dx=cos4x4+xcos4x4sin4x16+C\int(1+x) \sin (4 x) d x=-\frac{\cos 4 x}{4}+\frac{x \cos 4 x}{4}-\frac{\sin 4 x}{16}+C

Question 20

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(1+x)sin(4x)dx=cos4x4+xcos4x4sin4x16+C\int(1+x) \sin (4 x) d x=-\frac{\cos 4 x}{4}+\frac{x \cos 4 x}{4}-\frac{\sin 4 x}{16}+C .

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