∫32x3(log x)2dx is equal to

Question

∫32x3(log x)2dx is equal to (A)  8x4(log x)2 + C  (B)  x4 8(log x)2 - 4log x + 1 + C (C)  x4 8(log x)2 - 4log x + C (D)  x3 (log x)2 +2log x + C

Tardigrade Admin · Accepted Answer

∫32x3(log x)2dx 32 (log x2) (x4/4)-∫2log x ⋅(1/x)⋅(x4/4)dx +C' =8x4(logx)2-16∫ x3log x dx+C' = 8x4(log x)2-16 logx ⋅(x4/4)-∫(1/x)⋅(x4/4)dx +C'' = 8x4(log x)2-4x4logx +4 ∫ x3dx +C'' =8x4(logx)2 +4x4logx+x4+C = x4 [8(logx)2-4logx +1]

Q. $\int 32 x^{3} (l o gx)^{2} d x$ is equal to

$8 x^{4} (l o gx)^{2} + C$

$x^{4} {8 (l o gx)^{2} - 4 l o gx + 1} + C$

$x^{4} {8 (l o gx)^{2} - 4 l o gx} + C$

$x^{3} {(l o gx)^{2} + 2 l o gx} + C$

Q. ∫32x3(logx)2dx is equal to

8x4(logx)2+C

x4{8(logx)2−4logx+1}+C

x4{8(logx)2−4logx}+C

x3{(logx)2+2logx}+C

∫32x3(logx)2dx 32{(logx2)4x4​−∫2logx⋅x1​⋅4x4​dx}+C′ =8x4(logx)2−16∫x3logxdx+C′ =8x4(logx)2−16{logx⋅4x4​−∫x1​⋅4x4​dx}+C′′ =8x4(logx)2−4x4logx+4∫x3dx+C′′ =8x4(logx)2+4x4logx+x4+C =x4[8(logx)2−4logx+1]

Q. $\int 32 x^{3} (l o gx)^{2} d x$ is equal to

$8 x^{4} (l o gx)^{2} + C$

$x^{4} {8 (l o gx)^{2} - 4 l o gx + 1} + C$

$x^{4} {8 (l o gx)^{2} - 4 l o gx} + C$

$x^{3} {(l o gx)^{2} + 2 l o gx} + C$