Let \(I=\int \frac{e^{6 \log x}-e^{5 \log x}}{e^{4 \log x}-e^{3 \log x}} d x=\int \frac{e^{\log x^{\circ}}-e^{\log x^{3}}}{e^{\log x^{4}}-e^{\log x^{3}}} d x\) 
\(=\int \frac{x^{6}-x^{5}}{x^{4}-x^{3}} d x \quad\left[\because e^{\log f(x)}=f(x)\right][1]\)
\(=\int \frac{x^{5}(x-1)}{x^{3}(x-1)} d x=\int x^{2} d x=\frac{x^{3}}{3}+C \) 

\(\text { Use the formula, } 1+\sin 2 x=(\cos x+\sin x)^{2} \text { . }\)

On putting 1+x² = t, given integral reduces to \(\frac{1}{2} \int_{1}^{2} \frac{d t}{\sqrt{t}}\) 
=\(\sqrt{2}-1\)

On putting sin x = t, given integral reduces to
\(\int_{0}^{1} e^{t} d t .[\text { Ans. } e-1]\)

Let \(I=\int_{0}^{1} \frac{d x}{e^{x}+e^{-x}}=\int_{0}^{1} \frac{e^{x}}{e^{2 x}+1} d x\) 
Now, put \(e^{x}=t \Rightarrow e^{x} d x=d t\) 
Upper limit When x = 1, then t = e¹ = e
Lower limit When x = 0, then t = e^o = 1
\(\therefore \ I=\int_{1}^{e} \frac{d t}{t^{2}+1}=\left[\tan ^{-1} t\right]_{1}^{e}\) 
\(=\tan ^{-1} e-\tan ^{-1} 1=\tan ^{-1} e-\frac{\pi}{4}\)