1/x/arctan(frac{sin(x+1){ln(x+3x+6)})} सोडवा

$$\frac { \frac { 1 } { x } } { \arctan ( \frac { \sin ( x + 1 ) } { \ln ( x + 3 x + 6 ) } ) }$$

मूल्यांकन करा

$$\frac{1}{x\arctan(\frac{\sin(x+1)}{\ln(2\left(2x+3\right))})}$$

x संदर्भात फरक करा

$$\frac{-4x\left(\sin(1)\cos(x)\right)^{2}\arctan(\frac{\sin(x+1)}{\ln(2\left(2x+3\right))})-4x\left(\cos(1)\sin(x)\right)^{2}\arctan(\frac{\sin(x+1)}{\ln(2\left(2x+3\right))})-4x\ln(2\left(2x+3\right))^{2}\arctan(\frac{\sin(x+1)}{\ln(2\left(2x+3\right))})-6\left(\sin(1)\cos(x)\right)^{2}\arctan(\frac{\sin(x+1)}{\ln(2\left(2x+3\right))})-6\left(\cos(1)\sin(x)\right)^{2}\arctan(\frac{\sin(x+1)}{\ln(2\left(2x+3\right))})-6\ln(2\left(2x+3\right))^{2}\arctan(\frac{\sin(x+1)}{\ln(2\left(2x+3\right))})-2\sin(2)x\sin(2x)\arctan(\frac{\sin(x+1)}{\ln(2\left(2x+3\right))})-4x^{2}\ln(2\left(2x+3\right))\cos(x+1)-3\sin(2)\sin(2x)\arctan(\frac{\sin(x+1)}{\ln(2\left(2x+3\right))})-6x\ln(2\left(2x+3\right))\cos(x+1)+4x\sin(x+1)}{\left(2x+3\right)\left(2\left(\sin(1)\cos(x)\right)^{2}+2\left(\cos(1)\sin(x)\right)^{2}+2\ln(2\left(2x+3\right))^{2}+\sin(2)\sin(2x)\right)\left(x\arctan(\frac{\sin(x+1)}{\ln(2\left(2x+3\right))})\right)^{2}}$$