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Problem For x > 1 x>1 x > 1 d d x [ ( 2 x + 1 ) x ( x 2 − 1 ) x ] \dfrac{d}{dx}\bigl[(2x+1)^{\sqrt{x}}(x^{2}-1)^{x}\bigr] d x d [ ( 2 x + 1 ) x ( x 2 − 1 ) x ] ( 2 x + 1 ) x ( x 2 − 1 ) x [ 2 x 2 x + 1 + 2 x 2 x 2 − 1 ] (2x+1)^{\sqrt{x}}(x^{2}-1)^{x}\left[\dfrac{2\sqrt{x}}{2x+1}+\dfrac{2x^{2}}{x^{2}-1}\right] ( 2 x + 1 ) x ( x 2 − 1 ) x [ 2 x + 1 2 x + x 2 − 1 2 x 2 ] ( 2 x + 1 ) x ( x 2 − 1 ) x [ ln ( 2 x + 1 ) 2 x + x 2 x + 1 + ln ( x 2 − 1 ) + 2 x 2 x 2 − 1 ] (2x+1)^{\sqrt{x}}(x^{2}-1)^{x}\left[\dfrac{\ln(2x+1)}{2\sqrt{x}}+\dfrac{\sqrt{x}}{2x+1}+\ln(x^{2}-1)+\dfrac{2x^{2}}{x^{2}-1}\right] ( 2 x + 1 ) x ( x 2 − 1 ) x [ 2 x ln ( 2 x + 1 ) + 2 x + 1 x + ln ( x 2 − 1 ) + x 2 − 1 2 x 2 ] ( 2 x + 1 ) x ( x 2 − 1 ) x [ ln ( 2 x + 1 ) 2 x + 2 x 2 x + 1 + ln ( x 2 − 1 ) + 2 x 2 x 2 − 1 ] (2x+1)^{\sqrt{x}}(x^{2}-1)^{x}\left[\dfrac{\ln(2x+1)}{2\sqrt{x}} + \dfrac{2\sqrt{x}}{2x+1} + \ln(x^{2}-1) + \dfrac{2x^{2}}{x^{2}-1}\right] ( 2 x + 1 ) x ( x 2 − 1 ) x [ 2 x ln ( 2 x + 1 ) + 2 x + 1 2 x + ln ( x 2 − 1 ) + x 2 − 1 2 x 2 ] ( 2 x + 1 ) x ( x 2 − 1 ) x [ ln ( 2 x + 1 ) 2 x + 2 x 2 x + 1 + ln ( x 2 − 1 ) + 2 x x 2 − 1 ] (2x+1)^{\sqrt{x}}(x^{2}-1)^{x}\left[\dfrac{\ln(2x+1)}{2\sqrt{x}}+\dfrac{2\sqrt{x}}{2x+1}+\ln(x^{2}-1)+\dfrac{2x}{x^{2}-1}\right] ( 2 x + 1 ) x ( x 2 − 1 ) x [ 2 x ln ( 2 x + 1 ) + 2 x + 1 2 x + ln ( x 2 − 1 ) + x 2 − 1 2 x ] Browse the problem library