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∫sen (3x)dx\int sen\ \left(3x\right)dx∫sen (3x)dx Al resolver la integral. obtenemos
−13cos(3x)+C-\frac{1}{3}\cos\left(3x\right)+C−31cos(3x)+C
13cos(3x)+C\frac{1}{3}\cos\left(3x\right)+C31cos(3x)+C
−13sen(3x)+C-\frac{1}{3}sen\left(3x\right)+C−31sen(3x)+C
13sen(3x)+C\frac{1}{3}sen\left(3x\right)+C31sen(3x)+C
∫ex(x2−2x−1)dx\int e^x\left(x^2-2x-1\right)dx∫ex(x2−2x−1)dx
ex(x2−4x+3)2+Ce^x\left(x^2-4x+3\right)^2+Cex(x2−4x+3)2+C
ex(x2+2x−1)+Ce^x\left(x^2+2x-1\right)+Cex(x2+2x−1)+C
ex(x2−4x+3)3+Ce^x\left(x^2-4x+3\right)^3+Cex(x2−4x+3)3+C
ex(u)2+Ce^x\left(u\right)^2+Cex(u)2+C
∫3x4x2+5dx\int\frac{3x}{4x^2+5}dx∫4x2+53xdx El resultado es
38ln∣4x2+5∣+C\frac{3}{8}\ln\left|4x^2+5\right|+C83ln∣∣4x2+5∣∣+C
−38ln∣4x2+5∣+C-\frac{3}{8}\ln\left|4x^2+5\right|+C−83ln∣∣4x2+5∣∣+C
83ln∣4x2+5∣+C\frac{8}{3}\ln\left|4x^2+5\right|+C38ln∣∣4x2+5∣∣+C
−83ln∣4x2+5∣+C-\frac{8}{3}\ln\left|4x^2+5\right|+C−38ln∣∣4x2+5∣∣+C
∫xx+1dx\int x\sqrt{x+1}dx∫xx+1dx Al resolver la integral, el resultado es
25(x+1)5+23(x+1)3+C\frac{2}{5}\sqrt{\left(x+1\right)^5}+\frac{2}{3}\sqrt{\left(x+1\right)^3}+C52(x+1)5+32(x+1)3+C
−25(x+1)5−23(x+1)3+C-\frac{2}{5}\sqrt{\left(x+1\right)^5}-\frac{2}{3}\sqrt{\left(x+1\right)^3}+C−52(x+1)5−32(x+1)3+C
25(x+1)5−23(x+1)3+C\frac{2}{5}\sqrt{\left(x+1\right)^5}-\frac{2}{3}\sqrt{\left(x+1\right)^3}+C52(x+1)5−32(x+1)3+C
−25(x+1)5+23(x+1)3+C-\frac{2}{5}\sqrt{\left(x+1\right)^5}+\frac{2}{3}\sqrt{\left(x+1\right)^3}+C−52(x+1)5+32(x+1)3+C
∫(3x2−2x+1)5(1−3x)dx\int\left(3x^2-2x+1\right)^5\left(1-3x\right)dx∫(3x2−2x+1)5(1−3x)dx El resultado es
−112(3x2−2x+1)6+C-\frac{1}{12}\left(3x^2-2x+1\right)^6+C−121(3x2−2x+1)6+C
112(3x2−2x+1)6+C\frac{1}{12}\left(3x^2-2x+1\right)^6+C121(3x2−2x+1)6+C
−112(3x2+2x+1)6+C-\frac{1}{12}\left(3x^2+2x+1\right)^6+C−121(3x2+2x+1)6+C
−112(3x2−2x−1)6+C-\frac{1}{12}\left(3x^2-2x-1\right)^6+C−121(3x2−2x−1)6+C
∫8x32x4+6dx\int8x^3\sqrt{2x^4+6}dx∫8x32x4+6dx La integral es
−2(2x4+6)33+C-\frac{2\sqrt{\left(2x^4+6\right)^3}}{3}+C−32(2x4+6)3+C
2(2x4+6)33+C\frac{2\sqrt{\left(2x^4+6\right)^3}}{3}+C32(2x4+6)3+C
2(2x4−6)33+C\frac{2\sqrt{\left(2x^4-6\right)^3}}{3}+C32(2x4−6)3+C
(2x4+6)33+C\frac{\sqrt{\left(2x^4+6\right)^3}}{3}+C3(2x4+6)3+C
∫(x(x2+3)5dx)\int\left(\frac{x}{\left(x^2+3\right)^5}dx\right)∫((x2+3)5xdx)
−18(x2+3)4+C-\frac{1}{8\left(x^2+3\right)^4}+C−8(x2+3)41+C
18(x2−3)4+C\frac{1}{8\left(x^2-3\right)^4}+C8(x2−3)41+C
18(x2+3)4+C\frac{1}{8\left(x^2+3\right)^4}+C8(x2+3)41+C
−18(x2+3)4−C-\frac{1}{8\left(x^2+3\right)^4}-C−8(x2+3)41−C
∫x2−xdx\int x\sqrt{2-x}dx∫x2−xdx
−43u32+2u52+C-\frac{4}{3}u^{\frac{3}{2}}+2u^{\frac{5}{2}}+C−34u23+2u25+C
−43u32−2u52+C-\frac{4}{3}u^{\frac{3}{2}}-2u^{\frac{5}{2}}+C−34u23−2u25+C
43u32+2u52+C\frac{4}{3}u^{\frac{3}{2}}+2u^{\frac{5}{2}}+C34u23+2u25+C
43u32−2u52+C\frac{4}{3}u^{\frac{3}{2}}-2u^{\frac{5}{2}}+C34u23−2u25+C
∫(6x3−4x)(3x4−4x2)dx\int\left(6x^3-4x\right)\left(3x^4-4x^2\right)dx∫(6x3−4x)(3x4−4x2)dx El resultado es
(3x4+4x2)816+C\frac{\left(3x^4+4x^2\right)^8}{16}+C16(3x4+4x2)8+C
(3x4−4x2)84+C\frac{\left(3x^4-4x^2\right)^8}{4}+C4(3x4−4x2)8+C
(3x4−4x2)816+C\frac{\left(3x^4-4x^2\right)^8}{16}+C16(3x4−4x2)8+C
−(3x4−4x2)84+C-\frac{\left(3x^4-4x^2\right)^8}{4}+C−4(3x4−4x2)8+C
∫(x+2x2+4x)\int\left(\frac{x+2}{\sqrt{x^2+4x}}\right)∫(x2+4xx+2) La integral es
(x2+4x)2+C\left(x^2+4x\right)^2+C(x2+4x)2+C
(x2+4x)3+C\sqrt{\left(x^2+4x\right)^3}+C(x2+4x)3+C
(x2+4x)4+C\sqrt{\left(x^2+4x\right)^4}+C(x2+4x)4+C
(x2+4x)+C\sqrt{\left(x^2+4x\right)}+C(x2+4x)+C
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