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Problem Define P ( x ) = x 2 ∫ 0 x ln ( 1 + t ) d t P(x)=x^{2}\displaystyle\int_{0}^{x} \ln(1+t)\,dt\, P ( x ) = x 2 ∫ 0 x ln ( 1 + t ) d t x > − 1 x> -1 x > − 1 P ′ ′ ( x ) P''(x) P ′′ ( x ) 2 ∫ 0 x ln ( 1 + t ) d t + 2 x ln ( 1 + x ) + x 2 1 + x 2\displaystyle\int_{0}^{x}\!\ln(1+t)\,dt\; +\; 2x\ln(1+x)\; +\; \dfrac{x^{2}}{1+x} 2 ∫ 0 x ln ( 1 + t ) d t + 2 x ln ( 1 + x ) + 1 + x x 2 2 ∫ 0 x ln ( 1 + t ) d t + 4 x ln ( 1 + x ) 2\displaystyle\int_{0}^{x}\!\ln(1+t)\,dt\; +\; 4x\ln(1+x) 2 ∫ 0 x ln ( 1 + t ) d t + 4 x ln ( 1 + x ) 2 ∫ 0 x ln ( 1 + t ) d t + 4 x ln ( 1 + x ) + x 2 1 + x 2\displaystyle\int_{0}^{x}\!\ln(1+t)\,dt\; +\; 4x\ln(1+x)\; +\; \dfrac{x^{2}}{1+x} 2 ∫ 0 x ln ( 1 + t ) d t + 4 x ln ( 1 + x ) + 1 + x x 2 2 ∫ 0 x ln ( 1 + t ) d t + 4 x ln ( 1 + x ) + x 2\displaystyle\int_{0}^{x}\!\ln(1+t)\,dt\; +\; 4x\ln(1+x)\; +\; x 2 ∫ 0 x ln ( 1 + t ) d t + 4 x ln ( 1 + x ) + x Browse the problem library