66 bài tập Tìm nguyên hàm của một số hàm số thường gặp (dùng bảng nguyên hàm) (có lời giải) - Đề 1

\(\int {{x^6}} \;{\rm{d}}x = \frac{1}{7}{x^7} + C\)

\(\int {{x^2}} \;{\rm{d}}x = \frac{{{x^3}}}{3} + C\) trên \(\mathbb{R}\).

c) \(\int {\frac{1}{{\sqrt x }}} \;{\rm{d}}x = \int {{x^{ - \frac{1}{2}}}} \;{\rm{d}}x = 2{x^{\frac{1}{2}}} + C = 2\sqrt x + C\).

d) \(\int {{x^4}} \;{\rm{d}}x = \frac{{{x^5}}}{5} + C\);

e) \(\int {\frac{1}{{{x^3}}}} \;{\rm{d}}x = \int {\left( {{x^{ - 3}}} \right)} {\rm{d}}x = \frac{{{x^{ - 2}}}}{{ - 2}} + C =  - \frac{1}{{2{x^2}}} + C\);

f) \(\int {\sqrt x } \;{\rm{d}}x = \int {{x^{\frac{1}{2}}}} \;{\rm{d}}x = \frac{{{x^{\frac{3}{2}}}}}{3} = \frac{{2x\sqrt x }}{3} + C\).

Hàm só́ \(f(x)\) cân tìm là một nguyên hàm của hảm số \({f^\prime }(x) = 2x + \frac{1}{{{x^2}}}\).

Ta có: \(\int {\left( {2x + \frac{1}{{{x^2}}}} \right)} {\rm{d}}x = \int 2 x\;{\rm{d}}x + \int {\frac{1}{{{x^2}}}} \;{\rm{d}}x = {x^2} - \frac{1}{x} + C.\)

Do đó, hàm só́ \(f(x)\) có dạng \(f(x) = {x^2} - \frac{1}{x} + C,x \in (0; + \infty )\).

Theo già thiết, \(f(1) = {1^2} - 1 + C = 1\) nên \(C = 1\) và \(f(x) = {x^2} - \frac{1}{x} + 1,x \in (0; + \infty )\).

Đáp số: \(f(4) = \frac{{67}}{4}\).

\(\int {\frac{1}{{{{\sin }^2}x}}} \;{\rm{d}}x = - \cot x + C\) trên \((0;\pi )\).

\(\int 2 \sin \frac{x}{2}\cos \frac{x}{2}\;{\rm{d}}x = \int {\sin } x\;{\rm{d}}x =  - \cos x + C{\rm{. }}\)

\(\int {{3^x}} \;{\rm{d}}x = \frac{{{3^x}}}{{\ln 3}} + C\);

\(\int {{e^{2x}}} \;{\rm{d}}x = \int {{{\left( {{e^2}} \right)}^x}} \;{\rm{d}}x = \frac{{{e^{2x}}}}{{\ln {e^2}}} + C = \frac{{{e^{2x}}}}{2} + C\).

\(\int {\frac{{2\sin x}}{3}} \;{\rm{d}}x = \frac{2}{3}\int {\sin } x\;{\rm{d}}x =  - \frac{2}{3}\cos x + C\);          

\(\int {\frac{{{3^{x - 1}}}}{2}} \;{\rm{d}}x = \frac{1}{2}\int {\frac{{{3^x}}}{3}} \;{\rm{d}}x = \frac{1}{6}\int {{3^x}} \;{\rm{d}}x = \frac{{{3^x}}}{{6\ln 3}} + C\).

\(\int {\left( { - \frac{{\cos x}}{4}} \right)} {\rm{d}}x =  - \frac{1}{4}\int {\cos } x\;{\rm{d}}x =  - \frac{1}{4}\left( {\sin x + {C_1}} \right) =  - \frac{1}{4}\sin x + C\left( {C =  - \frac{{{C_1}}}{4}} \right)\);

\(\int {{2^{2x + 1}}} \;{\rm{d}}x = 2\int {{4^x}} \;{\rm{d}}x = 2\left( {\frac{{{4^x}}}{{\ln 4}} + {C_1}} \right) = \frac{{{4^x}}}{{\ln 2}} + C\left( {C = 2{C_1}} \right)\).

\(\int {\left( {3\cos x - \frac{4}{x}} \right)} {\rm{d}}x = 3\int {\cos } x\;{\rm{d}}x - 4\int {\frac{1}{x}} \;{\rm{d}}x = 3\sin x - 4\ln |x| + C(x \ne 0)\);

\(\int {{{(2x + 1)}^3}} \;{\rm{d}}x = \int {\left( {8{x^3} + 12{x^2} + 6x + 1} \right)} {\rm{d}}x = 8\int {{x^3}} \;{\rm{d}}x + 12\int {{x^2}} \;{\rm{d}}x + 6\int x \;{\rm{d}}x + \int 1 \;{\rm{d}}x\) \( = 2{x^4} + 4{x^3} + 3{x^2} + x + C\).

\(\int {\left( {3{x^3} + \frac{2}{{\sqrt[5]{{{x^3}}}}}} \right)} {\rm{d}}x = 3\int {{x^3}} \;{\rm{d}}x + 2\int {{x^{ - \frac{3}{5}}}} \;{\rm{d}}x = 3 \cdot \frac{{{x^4}}}{4} + 2 \cdot \frac{{{x^{\frac{2}{5}}}}}{{\frac{2}{5}}} + C = \frac{{3{x^4}}}{4} + 5\sqrt[5]{{{x^2}}} + C\);

\(\int {\left( {\frac{3}{{{{\cos }^2}x}} - \frac{1}{{{{\sin }^2}x}}} \right)} {\rm{d}}x = 3\int {\frac{1}{{{{\cos }^2}x}}} \;{\rm{d}}x - \int {\frac{1}{{{{\sin }^2}x}}} \;{\rm{d}}x = 3\tan x + \cot x + C\).

\(\int {{x^5}} \;{\rm{d}}x = \frac{{{x^6}}}{6} + C\);

\(\int {\frac{1}{{\sqrt[3]{{{x^2}}}}}} \;{\rm{d}}x = \int {{x^{ - \frac{2}{3}}}} \;{\rm{d}}x = 3{x^{\frac{1}{3}}} + C = 3\sqrt[3]{x} + C(x > 0)\);

\(\int {{7^x}} \;{\rm{d}}x = \frac{{{7^x}}}{{\ln 7}} + C\)

\(\int {\frac{{{3^x}}}{{{5^x}}}} \;{\rm{d}}x = \int {{{\left( {\frac{3}{5}} \right)}^x}} \;{\rm{d}}x = \frac{{{{\left( {\frac{3}{5}} \right)}^x}}}{{\ln \frac{3}{5}}} + C = \frac{{{3^x}}}{{(\ln 3 - \ln 5){5^x}}} + C\).

\(\int {\left( {2{x^5} + 3} \right)} {\rm{d}}x = 2\int {{x^5}} \;{\rm{d}}x + 3\int {\rm{d}} x = \frac{{{x^6}}}{3} + 3x + C\);

 \(\int {(5\cos x - 3\sin x)} {\rm{d}}x = 5\int {\cos } x\;{\rm{d}}x - 3\int {\sin } x\;{\rm{d}}x = 5\sin x + 3\cos x + C\);

\(\int {\left( {\frac{{\sqrt x }}{2} - \frac{2}{x}} \right)} {\rm{d}}x{\rm{ }} = \frac{1}{2}\int {{x^{\frac{1}{2}}}} \;{\rm{d}}x - 2\int {\frac{1}{x}} \;{\rm{d}}x = \frac{1}{2} \cdot \frac{1}{3} \cdot {x^{\frac{3}{2}}} - 2\ln x + C = \frac{1}{3}x\sqrt x  - 2\ln x + C(x > 0)\)

\(\int {\left( {{e^{x - 2}} - \frac{2}{{{{\sin }^2}x}}} \right)} {\rm{d}}x{\rm{ }} = \frac{1}{{{e^2}}}\int {{e^x}} \;{\rm{d}}x - 2\int {\frac{1}{{{{\sin }^2}x}}} \;{\rm{d}}x = \frac{{{e^x}}}{{{e^2}}} + 2\cot x + C = {e^{x - 2}} + 2\cot x + C\)

\(\int x {(2x - 3)^2}\;{\rm{d}}x = \int {\left( {4{x^3} - 12{x^2} + 9x} \right)} {\rm{d}}x\)\( = 4\int {{x^3}} \;{\rm{d}}x - 12\int {{x^2}} \;{\rm{d}}x + 9\int x \;{\rm{d}}x = {x^4} - 4{x^3} + \frac{{9{x^2}}}{2} + C\)

\(\int {{{\sin }^2}} \frac{x}{2}\;{\rm{d}}x = \int {\frac{{1 - \cos x}}{2}} \;{\rm{d}}x = \frac{1}{2}\left( {\int {\rm{d}} x - \int {\cos } x\;{\rm{d}}x} \right) = \frac{1}{2}(x - \sin x) + C\)

\(\int {{{\tan }^2}} x\;{\rm{d}}x = \int {\left( {\frac{1}{{{{\cos }^2}x}} - 1} \right)} {\rm{d}}x = \int {\frac{1}{{{{\cos }^2}x}}} \;{\rm{d}}x - \int {\rm{d}} x = \tan x - x + C\)

\(\int {{2^{3x}}}  \cdot {3^x}\;{\rm{d}}x = \int 2 {4^x}\;{\rm{d}}x = \frac{{{{24}^x}}}{{\ln 24}} + C\)