bài tập nguyên hàm

1. Tính các nguyên hàm sau: $I_1=\int(\frac{4}{5}x^3-5x^2-2x-3)dx$ $=\frac{4}{5}\cdot\frac{x^4}{4}-5\cdot\frac{x^3}{3}-2\cdot\frac{x^2}{2}-3x+C$ $=\frac{x^4}{5}-\frac{5x^3}{3}-x^2-3x+C$ $I_2=\int(\frac{4}{3}x^4-4x^3-2x+1)dx$ $=\frac{4}{3}\cdot\frac{x^5}{5}-4\cdot\frac{x^4}{4}-2\cdot\frac{x^2}{2}+x+C$ $=\frac{4x^5}{15}-x^4-x^2+x+C$ $I_3=\int(\frac{2}{3}x^4+5x^2-3x-4)dx$ $=\frac{2}{3}\cdot\frac{x^5}{5}+5\cdot\frac{x^3}{3}-3\cdot\frac{x^2}{2}-4x+C$ $=\frac{2x^5}{15}+\frac{5x^3}{3}-\frac{3x^2}{2}-4x+C$ $I_4=\int(-\frac{1}{3}x^2+4x-1)dx$ $=-\frac{1}{3}\cdot\frac{x^3}{3}+4\cdot\frac{x^2}{2}-x+C$ $=-\frac{x^3}{9}+2x^2-x+C$ $I_5=\int(x^{\frac{1}{4}}-x^{-1}-2)dx$ $=\frac{x^{\frac{5}{4}}}{\frac{5}{4}}-\ln|x|-2x+C$ $=\frac{4x^{\frac{5}{4}}}{5}-\ln|x|-2x+C$ $I_6=\int(\frac{3}{2}x^{\frac{5}{4}}+4x^{\frac{1}{4}}-\frac{1}{3}x)dx$ $=\frac{3}{2}\cdot\frac{x^{\frac{9}{4}}}{\frac{9}{4}}+4\cdot\frac{x^{\frac{5}{4}}}{\frac{5}{4}}-\frac{1}{3}\cdot\frac{x^2}{2}+C$ $=\frac{2x^{\frac{9}{4}}}{3}+\frac{16x^{\frac{5}{4}}}{5}-\frac{x^2}{6}+C$ 2. Tính các nguyên hàm sau: $I_1=\int(2\sin2x-4\cos3x)dx$ $=2\cdot\frac{-\cos2x}{2}-4\cdot\frac{\sin3x}{3}+C$ $=-\cos2x-\frac{4\sin3x}{3}+C$ $I_2=\int(2\cos7x-\frac{1}{2}x^2)dx$ $=2\cdot\frac{\sin7x}{7}-\frac{1}{2}\cdot\frac{x^3}{3}+C$ $=\frac{2\sin7x}{7}-\frac{x^3}{6}+C$ $I_3=\int(2\sin\frac{x}{2}-3\cos2x)dx$ $=2\cdot\frac{-\cos\frac{x}{2}}{\frac{1}{2}}-3\cdot\frac{\sin2x}{2}+C$ $=-4\cos\frac{x}{2}-\frac{3\sin2x}{2}+C$ $I_4=\int(\frac{1}{2}\cos3x-4\sin\frac{x}{3})dx$ $=\frac{1}{2}\cdot\frac{\sin3x}{3}-4\cdot\frac{-\cos\frac{x}{3}}{\frac{1}{3}}+C$ $=\frac{\sin3x}{6}+12\cos\frac{x}{3}+C$ $I_5=\int(\frac{1}{2}\sin2x+2\cos4x)dx$ $=\frac{1}{2}\cdot\frac{-\cos2x}{2}+2\cdot\frac{\sin4x}{4}+C$ $=-\frac{\cos2x}{4}+\frac{\sin4x}{2}+C$ $I_6=\int(8\cos2x-\sin(2x+\frac{\pi}{4}))dx$ $=8\cdot\frac{\sin2x}{2}-\frac{-\cos(2x+\frac{\pi}{4})}{2}+C$ $=4\sin2x+\frac{\cos(2x+\frac{\pi}{4})}{2}+C$ 3. Tính các nguyên hàm sau: $I_1=\int(2x-\frac{2}{3x+1})dx$ $=2\cdot\frac{x^2}{2}-2\cdot\frac{\ln|3x+1|}{3}+C$ $=x^2-\frac{2\ln|3x+1|}{3}+C$ $I_2=\int(4x^3-\frac{3}{3-2x})dx$ $=4\cdot\frac{x^4}{4}-3\cdot\frac{\ln|3-2x|}{-2}+C$ $=x^4+\frac{3\ln|3-2x|}{2}+C$ $I_3=\int(\frac{1}{1-2x}+\frac{2}{4x+3})dx$ $=\frac{\ln|1-2x|}{-2}+2\cdot\frac{\ln|4x+3|}{4}+C$ $=-\frac{\ln|1-2x|}{2}+\frac{\ln|4x+3|}{2}+C$ $I_4=\int(e^{-2x}-\frac{1}{5x+3})dx$ $=\frac{e^{-2x}}{-2}-\frac{\ln|5x+3|}{5}+C$ $=-\frac{e^{-2x}}{2}-\frac{\ln|5x+3|}{5}+C$ $I_5=\int(e^{4x-1}-e^{-6x})dx$ $=\frac{e^{4x-1}}{4}-\frac{e^{-6x}}{-6}+C$ $=\frac{e^{4x-1}}{4}+\frac{e^{-6x}}{6}+C$ $I_6=\int(\frac{3}{x^2}-\frac{1}{x+3})dx$ $=3\cdot\frac{-1}{x}-\ln|x+3|+C$ $=-\frac{3}{x}-\ln|x+3|+C$ 4. Cho hàm số $f(x)=3x^2-4x+2.$ Tìm nguyên hàm $F(x)$ của hàm số $f(x)$ biết $F(2)=3.$ $F(x)=\int(3x^2-4x+2)dx$ $=3\cdot\frac{x^3}{3}-4\cdot\frac{x^2}{2}+2x+C$ $=x^3-2x^2+2x+C$ Biết $F(2)=3$, thay vào ta có: $2^3-2\cdot2^2+2\cdot2+C=3$ $8-8+4+C=3$ $C=-1$ Vậy $F(x)=x^3-2x^2+2x-1$ 5. Cho hàm số $f(x)=3\sin2x.$ Tìm nguyên hàm $F(x)$ của hàm số $f(x)$ biết $F(\frac{\pi}{6})=1.$ $F(x)=\int3\sin2xdx$ $=3\cdot\frac{-\cos2x}{2}+C$ $=-\frac{3\cos2x}{2}+C$ Biết $F(\frac{\pi}{6})=1$, thay vào ta có: $-\frac{3\cos(\frac{\pi}{3})}{2}+C=1$ $-\frac{3\cdot\frac{1}{2}}{2}+C=1$ $-\frac{3}{4}+C=1$ $C=\frac{7}{4}$ Vậy $F(x)=-\frac{3\cos2x}{2}+\frac{7}{4}$ 6. Cho hàm số $f(x)=e^{2x}.$ Tìm nguyên hàm $F(x)$ của hàm số $f(x)$ biết $F(\ln2)=5.$ $F(x)=\int e^{2x}dx$ $=\frac{e^{2x}}{2}+C$ Biết $F(\ln2)=5$, thay vào ta có: $\frac{e^{2\ln2}}{2}+C=5$ $\frac{e^{\ln4}}{2}+C=5$ $\frac{4}{2}+C=5$ $2+C=5$ $C=3$ Vậy $F(x)=\frac{e^{2x}}{2}+3$ 7. Cho hàm số $f(x)=\cos(x+\frac{\pi}{4})+3.$ Tìm nguyên hàm $F(x)$ của hàm số $f(x)$ biết $F(0)=2\sqrt{2}.$ $F(x)=\int(\cos(x+\frac{\pi}{4})+3)dx$ $=\sin(x+\frac{\pi}{4})+3x+C$ Biết $F(0)=2\sqrt{2}$, thay vào ta có: $\sin(0+\frac{\pi}{4})+3\cdot0+C=2\sqrt{2}$ $\sin\frac{\pi}{4}+C=2\sqrt{2}$ $\frac{\sqrt{2}}{2}+C=2\sqrt{2}$ $C=\frac{3\sqrt{2}}{2}$ Vậy $F(x)=\sin(x+\frac{\pi}{4})+3x+\frac{3\sqrt{2}}{2}$ 8. Cho hàm số $f(x)=\sin(\frac{3}{2}x).$ Tìm nguyên hàm $F(x)$ của hàm số $f(x)$ biết $F(\frac{\pi}{9})=3.$ $F(x)=\int\sin(\frac{3}{2}x)dx$ $=-\frac{\cos(\frac{3}{2}x)}{\frac{3}{2}}+C$ $=-\frac{2\cos(\frac{3}{2}x)}{3}+C$ Biết $F(\frac{\pi}{9})=3$, thay vào ta có: $-\frac{2\cos(\frac{3}{2}\cdot\frac{\pi}{9})}{3}+C=3$ $-\frac{2\cos(\frac{\pi}{6})}{3}+C=3$ $-\frac{2\cdot\frac{\sqrt{3}}{2}}{3}+C=3$ $-\frac{\sqrt{3}}{3}+C=3$ $C=3+\frac{\sqrt{3}}{3}$ Vậy $F(x)=-\frac{2\cos(\frac{3}{2}x)}{3}+3+\frac{\sqrt{3}}{3}$ 10. Tính các nguyên hàm sau: $I_1=\int\sin^2xdx$ $=\int\frac{1-\cos2x}{2}dx$ $=\frac{1}{2}\int dx-\frac{1}{2}\int\cos2xdx$ $=\frac{x}{2}-\frac{\sin2x}{4}+C$ $I_2=\int\cos^3xdx$ $=\int\cos^2x\cdot\cos xdx$ $=\int(1-\sin^2x)\cdot\cos xdx$ $=\int\cos xdx-\int\sin^2x\cdot\cos xdx$ $=\sin x-\frac{\sin^3x}{3}+C$ $I_3=\int\tan^2xdx$ $=\int(\sec^2x-1)dx$ $=\tan x-x+C$ $I_4=\int\cot^2xdx$ $=\int(\csc^2x-1)dx$ $=-\cot x-x+C$ 11. Tính các nguyên hàm sau: $I_1=\int\frac{x^2-3x+4}{x-1}dx$ $=\int(x-2+\frac{2}{x-1})dx$ $=\frac{x^2}{2}-2x+2\ln|x-1|+C$ $I_2=\int\frac{2x^2-4x+1}{x+2}dx$ $=\int(2x-8+\frac{17}{x+2})dx$ $=x^2-8x+17\ln|x+2|+C$ $I_3=\int\frac{x^3-2x^2+3}{x+1}dx$ $=\int(x^2-3x+3+\frac{0}{x+1})dx$ $=\frac{x^3}{3}-\frac{3x^2}{2}+3x+C$ 12. Tính các nguyên hàm sau: $I_1=\int\frac{1}{x^2-1}dx$ $=\int\frac{1}{(x-1)(x+1)}dx$ $=\frac{1}{2}\int(\frac{1}{x-1}-\frac{1}{x+1})dx$ $=\frac{1}{2}(\ln|x-1|-\ln|x+1|)+C$ $=\frac{1}{2}\ln|\frac{x-1}{x+1}|+C$ $I_2=\int\frac{2}{3x^2+x-2}dx$ $=\int\frac{2}{(3x-2)(x+1)}dx$ $=\frac{2}{5}\int(\frac{1}{x+1}-\frac{3}{3x-2})dx$ $=\frac{2}{5}(\ln|x+1|-\ln|3x-2|)+C$ $=\frac{2}{5}\ln|\frac{x+1}{3x-2}|+C$ $I_3=\int\frac{2x+1}{x^2-7x+10}dx$ $=\int\frac{2x+1}{(x-2)(x-5)}dx$ $=\frac{1}{3}\int(\frac{1}{x-2}+\frac{5}{x-5})dx$ $=\frac{1}{3}(\ln|x-2|+\ln|x-5|)+C$ $=\frac{1}{3}\ln|(x-2)(x-5)|+C$