Integrales Indefinidas

RESOLVER LAS INTEGRALES INDEFINIDAS

∫▒√(1/t-1).dt/t^2

∫▒√(1/t-1).dt/t^2 =-∫▒√(t^(-1)-1) ((-dt)/t^2 )=-∫▒〖√y dy〗=-2/3 [y^(3/2) ]+C=-2/3 (1/t-1)^(3/2)+C

∫▒dy/(√y+1)

∫▒dy/(√y+1)=∫▒(2u du)/(〖〖(u〗^2)〗^(1/2)+1)=2∫▒〖u/(u+1) du〗=2∫▒(-1+1/(u+1))du =2∫▒〖-1 du〗+2∫▒1/(u+1) du=2[-u+ln⁡(u+1) ]+C=-2√y+2 ln⁡(√y+1)+C

∫▒〖(〖x^3+3)〗^(1/4).x^5 〗 dx

∫▒〖(〖x^3+3)〗^(1/4).x^5 〗 dx=1/3 ∫▒〖〖〖(x〗^3+5)〗^(1/4) x^3 (3x^2 dx) 〗=1/3 ∫▒〖y^(1/4) (y-5) dy〗=1/3 ∫▒y^(5/4) +1/3 ∫▒〖5y^(1/4) 〗=1/3 ∫▒y^(5/4) +5/3 ∫▒y^(1/4) =1/3 (4/9)[y^(9/4) ]+5/3 (4/5)[y^(5/4) ]+C=4/27 [y^(9/4) ]+4/3 [y^(5/4) ]+C=4/27 〖(x^3+5)〗^(9/4)+4/3 〖(x^3+5)〗^(5/4)+C

∫▒sin⁡(1/3 x)dx

∫▒sin⁡(1/3 x)dx =3∫▒sin⁡(1/3 x) (1/3)dx=3∫▒sin⁡y dy=3(〖-cos〗⁡y )+C=-3[cos⁡y ]+C=-3 cos⁡〖1/3 x〗+C

∫▒〖1/2 t.cos⁡〖(4t^2)dt〗 〗

∫▒〖1/2 t.cos⁡〖(4t^2)dt〗 〗=1/16 ∫▒〖8t cos⁡〖(4t^2)〗 dt〗=1/16 ∫▒cos⁡(4t^2 ) (8t dt)=1/16 ∫▒cos⁡y dy=1/16 (sin⁡y )+C=1/16 sin⁡〖4t^2 〗+C

∫▒〖x^2 e^(-x^3 ) 〗 dx

∫▒〖x^2 e^(-x^3 ) 〗 dx=∫▒x^2/e^(-x^3 ) dx=1/3 ∫▒1/e^(-x^3 ) (3x^2 dx)=1/3 ∫▒〖1/e^y dy=1/3 ∫▒e^(-y) 〗 dy=1/3 (-∫▒〖e^y (-dy) 〗)=-1/3 ∫▒〖e^z dz〗= -1/3 [e^z ]+C=-1/3 [e^(-y) ]+c=-1/3 [e^(-x^3 ) ]+C=-1/(3e^(x^3 ) )+C

∫▒〖tan⁡x/cos^2⁡x dx〗

∫▒〖tan⁡x/cos^2⁡x dx〗=∫▒〖tan⁡x sec^2⁡x 〗 dx=tan⁡x 〖(tan〗⁡x)-∫▒〖tan⁡x sec^2⁡x 〗 dx=1/2 tan^2⁡x

∫▒〖tan⁡x sec^2⁡x 〗 dx=tan⁡x 〖(tan〗⁡x)-∫▒〖tan⁡x sec^2⁡x 〗 dx

2∫▒〖tan⁡x sec^2⁡x 〗 dx=tan^2⁡x

∫▒〖tan⁡x sec^2⁡x 〗 dx=1/2 tan^2⁡x

∫▒e^(x ln⁡a ) e^x dx ; u=x(ln⁡a+1)

∫▒e^(x ln⁡a ) e^x dx= ∫▒〖e^(x+x ln⁡a ) dx〗=∫▒〖e^(x 〖(ln〗⁡〖a+1〗)) dx〗=∫▒〖e^(x 〖(ln〗⁡〖a+1〗)) dx〗.ln⁡〖a+1〗/ln⁡〖a+1〗 =∫▒1/(〖(ln〗⁡〖a+1〗)) e^u du=1/(〖(ln〗⁡〖a+1〗)) ∫▒〖e^u du〗=1/(〖(ln〗⁡〖a+1〗)) [e^u ]+C=1/(〖(ln〗⁡〖a+1〗)) e^(x 〖(ln〗⁡〖a+1〗))+C

∫▒(x^2+1)/(x-1) dx=∫▒〖(〖(x〗^2-1)+2)/(x-1) dx〗

∫▒(x^2+1)/(x-1) dx=∫▒〖(〖(x〗^2-1)+2)/(x-1) dx〗=∫▒〖((x-1)(x+1)+2)/(x-1) dx=〗 ∫▒〖[(x-1)(x+1)/(x-1)+2/(x-1)]dx=〗 ∫▒(x+1+2/(x-1))dx=∫▒〖x dx〗+∫▒dx+2∫▒〖1/(x-1) dx〗=1/2 x^2+x+2/ln⁡|x-1| +C

∫▒〖dx/(2^x+3)=1/3 ∫▒((2^x+3)-2^x)/(2^x+3)〗 dx ;u=2^x+3

∫▒〖dx/(2^x+3)=1/3 ∫▒((2^x+3)-2^x)/(2^x+3)〗 dx=1/3 ∫▒((2^x+3)-2^x)/(2^x+3) ((2^x ln⁡2)/(2^x ln⁡2 ))dx=1/3 ∫▒(u-(u-3))/u (du/((u-3) ln⁡2 ))=1/3 ∫▒(3 du)/(u(u-3) ln⁡2 )=1/ln⁡2 ∫▒du/u(u-3) =1/ln⁡2 [u ln⁡u-3 ln⁡〖u-∫▒ln⁡〖u du〗 〗 ]=1/ln⁡2 [u ln⁡u-3 ln⁡〖u-〗 u ln⁡u+u]+C=1/ln⁡2 [u-3 ln⁡u ]+C=2^x/ln⁡2 +3/ln⁡2 -3 ln⁡|2^x+3|+C

∫▒ln⁡〖u du〗 = u ln⁡u-u+C

∫▒〖x/(1+√x) dx 〗 ;u=1+√x

∫▒〖x/(1+√x) dx 〗=2∫▒〖(x√x)/(1+√x) (dx/(2√x))=〗 2∫▒〖(u-1)^3/u=2∫▒〖(u^3-3u^2+3u-1)/u du=2∫▒〖(u^2-3u+3〗-1/u) du〗〗=2∫▒〖u^2 du-6∫▒〖u du+6∫▒〖du-2∫▒〖1/u du〗〗〗〗={2/3 [u^3 ]-3u^2+6u-2 ln⁡|u| }+C=2/3 〖(1+√x)〗^3-3(1+√x)^2+6(1+√x)-2 ln⁡〖(u)〗

∫▒〖csc^2⁡〖x dx〗=〗 ∫▒〖1/sin^2⁡x dx=∫▒〖cos^2⁡x/(sin^2⁡x cos^2⁡x ) dx〗 〗 ;u=tan⁡x

∫▒〖csc^2⁡〖x dx〗=〗 ∫▒〖1/sin^2⁡x dx=∫▒〖cos^2⁡x/(sin^2⁡x cos^2⁡x ) dx〗= 〗 ∫▒sec^2⁡x/tan^2⁡x dx=∫▒〖du/u^2 =〗 [u^(-1)/(-1)]+C=[-1/u]+C=-1/tan⁡x +C=-cot⁡〖x+C〗

∫▒〖y.csc⁡〖(3y^2)cot⁡〖(〖3y〗^2 ) dy〗 〗 〗

∫▒〖y.csc⁡〖(3y^2)cot⁡〖(〖3y〗^2 ) dy〗 〗 〗=1/6 ∫▒csc⁡〖(3y^2)cot⁡(〖3y〗^2 )(6y dy) 〗 =1/6 ∫▒〖〖(csc〗⁡z cot⁡〖z) dz〗 〗=1/6 [-csc⁡z ]+C=-1/6 csc⁡(3y^2 )+C

∫▒〖cos⁡x (2+sin⁡x )^5.dx〗

∫▒〖cos⁡x (2+sin⁡x )^5.dx=〗 1/5 ∫▒(2+sin⁡x )^5/(2+sin⁡x )^4 [5(2+sin⁡x )^5 cos⁡〖x dx〗 ]=1/5 ∫▒〖u/〖(2+√(5&u)-2)〗^4 du〗=1/5 ∫▒〖u/√(5&u^4 ) du〗=1/5 ∫▒〖u^(1/(5 )) du〗=1/5 (5/6)[u^(6/5) ]+C=1/6 〖((2+sin⁡x )^5)〗^(6/5)+C=1/6 (2+sin⁡x )^6+C

∫▒√(1+1/3x).dx/x^2

∫▒√(1+1/3x).dx/x^2 =-3∫▒√(1+1/3x).(-dx/〖3x〗^2 )=-3∫▒√(u ) du=-3(2/3)[u^(3/2) ]+C=-2[u^(3/2) ]+C=-2(1+1/3x)^(3/2)+C

∫▒〖2 sin⁡x ∛(1+cos⁡x ).dx〗

∫▒〖2 sin⁡x ∛(1+cos⁡x ).dx〗=-6∫▒〖∛(〖1+cos〗⁡x ) (1+cos⁡x )^(2/3) (-1/3 sin⁡x (1+cos⁡x )^(-2/3).dx) 〗=-6∫▒〖u(u^3 )^(2/3) 〗 du=-6∫▒〖u(u^2 ).du〗=-6∫▒〖u^3.du〗=-3/2

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