Dear student Let I=∫1xx5+3dx=∫x4x5x5+3dxPut x5+3=t⇒5x4dx=dt⇒x4dx=dt5So,I=15∫dttt-3=15∫dtt2-3t=15∫dtt2-3t+322-322=15∫dtt-322-322=15×12logt-32-32t-32+32+C=110logt-3t+C=110logx5+3-3x5+3+C=110logx5x5+3+C Regards

Integrate 1/x(x^5+3)

Dear student

Let I = \int \frac{1}{x (x^{5} + 3)} dx = \int \frac{x^{4}}{x^{5} (x^{5} + 3)} dx Put x^{5} + 3 = t \Rightarrow 5 x^{4} dx = dt \Rightarrow x^{4} dx = \frac{dt}{5} So, I = \frac{1}{5} \int \frac{dt}{t (t - 3)} = \frac{1}{5} \int \frac{dt}{t^{2} - 3 t} = \frac{1}{5} \int \frac{dt}{t^{2} - 3 t + {(\frac{3}{2})}^{2} - {(\frac{3}{2})}^{2}} = \frac{1}{5} \int \frac{dt}{{(t - \frac{3}{2})}^{2} - {(\frac{3}{2})}^{2}} = \frac{1}{5} \times \frac{1}{2} \log |\frac{t - \frac{3}{2} - \frac{3}{2}}{t - \frac{3}{2} + \frac{3}{2}}| + C = \frac{1}{10} \log |\frac{t - 3}{t}| + C = \frac{1}{10} \log |\frac{x^{5} + 3 - 3}{x^{5} + 3}| + C = \frac{1}{10} \log |\frac{x^{5}}{x^{5} + 3}| + C

Regards

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