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Question Number 219554    Answers: 0   Comments: 0

I_n =∫_0 ^( 1) ∫_0 ^1 ...∫_0 ^( 1) (((x_1 x_2 ...x_n )^a )/((1−x_1 x_2 ...x_n )))ln(x_(1 ) )ln(x_2 )...ln(x_(n ) )dx_1 dx_2 ...dx_(n )

In=0101...01(x1x2...xn)a(1x1x2...xn)ln(x1)ln(x2)...ln(xn)dx1dx2...dxn

Question Number 219553    Answers: 0   Comments: 0

I_n = ∫_0 ^( 1) ∫_0 ^( 1) ....∫_0 ^( 1) ((ln(1+x_1 x_2 ....x_n ))/((1−x_1 )(1−x_2 )....(1−x_(n ) ))) dx_1 dx_2 ....dx_n

In=0101....01ln(1+x1x2....xn)(1x1)(1x2)....(1xn)dx1dx2....dxn

Question Number 219552    Answers: 0   Comments: 0

Question Number 219550    Answers: 0   Comments: 0

Question Number 219549    Answers: 0   Comments: 0

Question Number 219529    Answers: 1   Comments: 1

∫_( 0) ^( 1) ((1+x−x^2 +x^3 −x^4 −x^5 )/(1−x^7 )) dx

011+xx2+x3x4x51x7dx

Question Number 219522    Answers: 0   Comments: 0

∫^( ∞) _( 0) ((10x^(17) e^(2x) (e^(2x) −1)+x^(17) e^x (e^(4x) −1))/((e^x −1)^6 )) dx

010x17e2x(e2x1)+x17ex(e4x1)(ex1)6dx

Question Number 219506    Answers: 2   Comments: 0

Q. Integrate ((x^2 +x+1)/(2x^2 −x−3)).

Q.Integratex2+x+12x2x3.

Question Number 219509    Answers: 2   Comments: 0

I = ∫^( 16) _( 1) (((x +(√x))^(1/4) )/x^(3/4) ) dx

I=116(x+x)14x34dx

Question Number 219482    Answers: 0   Comments: 0

∫ ((e^(−x^2 ) cos (x))/(x^2 +1)) dx

ex2cos(x)x2+1dx

Question Number 219458    Answers: 1   Comments: 0

∫_1 ^( 2) ∫_1 ^( 2) ∫_1 ^( 2) ∫_1 ^( 2) x_1 ^( 2) x_2 ^( 3) x_3 ^( 4) x_4 ^( 5) dx_1 dx_2 dx_3 dx_(4 )

12121212x12x23x34x45dx1dx2dx3dx4

Question Number 219456    Answers: 0   Comments: 2

∫_1 ^( 3) ∫_1 ^( 3) ∫_1 ^( 3) ∫_1 ^( 3) ∫_1 ^( 3) ((x_1 +x_2 +x_3 +x_4 −x_5 )/(x_1 +x_2 +x_3 +x_4 +x_5 )) dx_1 dx_2 dx_3 dx_4 dx_5

1313131313x1+x2+x3+x4x5x1+x2+x3+x4+x5dx1dx2dx3dx4dx5

Question Number 219453    Answers: 1   Comments: 0

∫_1 ^( 2) ∫_1 ^( 2) ∫_1 ^( 2) ∫_1 ^( 2) ((x_1 +x_(2 ) +x_3 −x_4 )/(x_1 +x_2 +x_3 +x_4 )) dx_1 dx_2 dx_3 dx_(4 )

12121212x1+x2+x3x4x1+x2+x3+x4dx1dx2dx3dx4

Question Number 219449    Answers: 0   Comments: 0

L{sinx}=∫_0 ^∞ e^(−sx) sinx dx=∫_0 ^∞ e^(−sx) ((e^(ix) −e^(−ix) )/(2i))dx =(1/(2i))[∫_0 ^∞ e^(−(s−i)x) dx −∫_0 ^∞ e^(−(s+i)x) dx] =(1/(2i))[((−1)/(s−i))e^(−(s−i)x) +(1/(s+i))e^(−(s+i)x) ]_0 ^∞ =(1/(2i))[(1/(s−i))−(1/(s+i))]=(1/(2i))×((s+i−s+i)/((s−i)(s+i)))=(1/(2i))×((2i)/(s^2 +1))=(1/(s^2 +1))

L{sinx}=0esxsinxdx=0esxeixeix2idx=12i[0e(si)xdx0e(s+i)xdx]=12i[1sie(si)x+1s+ie(s+i)x]0=12i[1si1s+i]=12i×s+is+i(si)(s+i)=12i×2is2+1=1s2+1

Question Number 219447    Answers: 0   Comments: 0

Lim_(x→∞) Σ_(i=1) ^∞ (((−x)/i))^i

Limxi=1(xi)i

Question Number 219446    Answers: 0   Comments: 0

∫tan(((1/n)/(sec(n)+(((1−sec(n))/(sec(n))))))dn

tan(1nsec(n)+(1sec(n)sec(n))dn

Question Number 219440    Answers: 0   Comments: 0

I=∫tan(((cos(n))/(n(1−cos(n) + cos^2 (n))))) dn

I=tan(cos(n)n(1cos(n)+cos2(n)))dn

Question Number 219434    Answers: 0   Comments: 0

Question Number 219428    Answers: 1   Comments: 1

a , b, ∈ R ∫_( −∞) ^( ∞) (((e^(iax) −1)(e^(ibx) −1))/x^2 ) dx

a,b,R(eiax1)(eibx1)x2dx

Question Number 219384    Answers: 1   Comments: 0

Question Number 219316    Answers: 1   Comments: 0

Prove; ∫^( ∞) _( 0) (1/(2^2^(⌊x⌋) + {x})) dx = ln2

Prove;0122x+{x}dx=ln2

Question Number 219305    Answers: 0   Comments: 0

Prove; I_0 (x) =(1/π)∫_0 ^( π) e^( x cox(θ)) dθ ; x^2 I_0 ^(′′) (x) + xI′_0 (x) − x^2 I_0 (x) = 0;

Prove;I0(x)=1π0πexcox(θ)dθ;x2I0(x)+xI0(x)x2I0(x)=0;

Question Number 219268    Answers: 1   Comments: 0

f(x,y)=ln∫_0 ^(x^2 +y^2 ) e^t^2 dt,f(x)(1,2)=?

f(x,y)=ln0x2+y2et2dt,f(x)(1,2)=?

Question Number 219236    Answers: 1   Comments: 1

f(t) = (1/(2πi)) ∫_( c−i∞) ^( c+i∞) (e^(st) /(s^k )) ds , k ∈C

f(t)=12πicic+iestskds,kC

Question Number 219234    Answers: 0   Comments: 0

f(t)=∫_(0 ) ^( t) ((ζ(1/2 + iτ))/( (√(t − τ + 1)))) dτ

f(t)=0tζ(1/2+iτ)tτ+1dτ

Question Number 219193    Answers: 2   Comments: 0

∫(√((x+1)/(x+2))) .(1/(x+3)) dx=?

x+1x+2.1x+3dx=?

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