\displaystyle \textbf{Evaluate the following integrals:}

\displaystyle \textbf{Question 1: }~\int \frac{1}{a^2-b^2x^2}\,dx.
\displaystyle \text{Answer:}

\displaystyle I=\int \frac{dx}{a^{2}-b^{2}x^{2}}

\displaystyle I=\frac{1}{b^{2}}\int \frac{dx}{\left(\frac{a^{2}}{b^{2}}-x^{2}\right)}

\displaystyle I=\frac{1}{b^{2}}\cdot\frac{1}{2\left(\frac{a}{b}\right)}\log\left|\frac{\frac{a}{b}+x}{\frac{a}{b}-x}\right|+C

\displaystyle I=\frac{1}{2ab}\log\left|\frac{a+bx}{a-bx}\right|+C

\displaystyle \textbf{Question 2: }~\int \frac{1}{a^2x^2-b^2}\,dx.
\displaystyle \text{Answer:}

\displaystyle I=\int \frac{dx}{a^{2}x^{2}-b^{2}}

\displaystyle I=\frac{1}{a^{2}}\int \frac{dx}{x^{2}-\left(\frac{b}{a}\right)^{2}}

\displaystyle I=\frac{1}{a^{2}}\cdot\frac{1}{2\left(\frac{b}{a}\right)}\log\left|\frac{x-\frac{b}{a}}{x+\frac{b}{a}}\right|+C

\displaystyle I=\frac{1}{2ab}\log\left|\frac{ax-b}{ax+b}\right|+C

\displaystyle \textbf{Question 3: }~\int \frac{1}{a^2x^2+b^2}\,dx.
\displaystyle \text{Answer:}

\displaystyle I=\int \frac{dx}{a^{2}x^{2}+b^{2}}

\displaystyle I=\frac{1}{a^{2}}\int \frac{dx}{x^{2}+\left(\frac{b}{a}\right)^{2}}

\displaystyle I=\frac{1}{a^{2}}\cdot\frac{a}{b}\tan^{-1}\left(\frac{x}{\frac{b}{a}}\right)+C

\displaystyle I=\frac{1}{ab}\tan^{-1}\left(\frac{ax}{b}\right)+C

\displaystyle \textbf{Question 4: }~\int \frac{x^2-1}{x^2+4}\,dx.
\displaystyle \text{Answer:}

\displaystyle I=\int \frac{x^{2}-1}{x^{2}+4}\,dx

\displaystyle I=\int \frac{(x^{2}+4)-5}{x^{2}+4}\,dx

\displaystyle I=\int \frac{x^{2}+4}{x^{2}+4}\,dx-5\int \frac{dx}{x^{2}+2^{2}}

\displaystyle I=\int dx-5\int \frac{dx}{x^{2}+2^{2}}

\displaystyle I=x-\frac{5}{2}\tan^{-1}\left(\frac{x}{2}\right)+C

\displaystyle \textbf{Question 5: }~\int \frac{1}{\sqrt{1+4x^2}}\,dx.
\displaystyle \text{Answer:}

\displaystyle I=\int \frac{dx}{\sqrt{1+4x^{2}}}

\displaystyle I=\int \frac{dx}{\sqrt{1+(2x)^{2}}}

\displaystyle \text{Let } 2x=t

\displaystyle \text{Then } 2\,dx=dt

\displaystyle dx=\frac{dt}{2}

\displaystyle I=\frac{1}{2}\int \frac{dt}{\sqrt{1+t^{2}}}

\displaystyle I=\frac{1}{2}\log\left|t+\sqrt{1+t^{2}}\right|+C

\displaystyle I=\frac{1}{2}\log\left|2x+\sqrt{1+4x^{2}}\right|+C

\displaystyle \textbf{Question 6: }~\int \frac{1}{\sqrt{a^2+b^2x^2}}\,dx.
\displaystyle \text{Answer:}

\displaystyle I=\int \frac{dx}{\sqrt{a^{2}+b^{2}x^{2}}}

\displaystyle =\int \frac{dx}{\sqrt{b^{2}\left(\frac{a^{2}}{b^{2}}+x^{2}\right)}}

\displaystyle =\frac{1}{b}\int \frac{dx}{\sqrt{x^{2}+\left(\frac{a}{b}\right)^{2}}}

\displaystyle =\frac{1}{b}\log\left|x+\sqrt{x^{2}+\frac{a^{2}}{b^{2}}}\right|+C

\displaystyle =\frac{1}{b}\left[\log\left|x+\frac{\sqrt{b^{2}x^{2}+a^{2}}}{b}\right|\right]+C

\displaystyle =\frac{1}{b}\left[\log\left|\frac{bx+\sqrt{b^{2}x^{2}+a^{2}}}{b}\right|\right]+C

\displaystyle =\frac{1}{b}\left[\log\left|bx+\sqrt{b^{2}x^{2}+a^{2}}\right|-\log b\right]+C

\displaystyle =\frac{1}{b}\log\left|bx+\sqrt{b^{2}x^{2}+a^{2}}\right|-\frac{\log b}{b}+C

\displaystyle \text{let } C-\frac{\log b}{b}=C'

\displaystyle =\frac{1}{b}\log\left|bx+\sqrt{b^{2}x^{2}+a^{2}}\right|+C'

\displaystyle \textbf{Question 7: }~\int \frac{1}{\sqrt{a^2-b^2x^2}}\,dx.
\displaystyle \text{Answer:}
\displaystyle I=\int \frac{dx}{\sqrt{a^{2}-b^{2}x^{2}}}

\displaystyle =\int \frac{dx}{\sqrt{b^{2}\left(\frac{a^{2}}{b^{2}}-x^{2}\right)}}

\displaystyle =\frac{1}{b}\int \frac{dx}{\sqrt{\left(\frac{a}{b}\right)^{2}-x^{2}}}

\displaystyle =\frac{1}{b}\sin^{-1}\left(\frac{bx}{a}\right)+C

\displaystyle \textbf{Question 8: }~\int \frac{1}{\sqrt{(2-x)^2+1}}\,dx.
\displaystyle \text{Answer:}

\displaystyle I=\int \frac{dx}{\sqrt{(2-x)^{2}+1}}

\displaystyle \text{Let } 2-x=t

\displaystyle \Rightarrow -dx=dt

\displaystyle \Rightarrow dx=-dt

\displaystyle \text{Now, } \int \frac{dx}{\sqrt{(2-x)^{2}+1}}

\displaystyle =-\int \frac{dt}{\sqrt{t^{2}+1}}

\displaystyle =-\log\left|t+\sqrt{t^{2}+1}\right|+C

\displaystyle =-\log\left|(2-x)+\sqrt{(2-x)^{2}+1}\right|+C

\displaystyle \textbf{Question 9: }~\int \frac{1}{\sqrt{(2-x)^2-1}}\,dx.
\displaystyle \text{Answer:}

\displaystyle I=\int \frac{dx}{\sqrt{(2-x)^{2}-1}}

\displaystyle \text{Let } 2-x=t

\displaystyle \Rightarrow -dx=dt

\displaystyle \Rightarrow dx=-dt

\displaystyle \text{Now, } \int \frac{dx}{\sqrt{(2-x)^{2}-1}}

\displaystyle =\int \frac{-dt}{\sqrt{t^{2}-1}}

\displaystyle =-\log\left|t+\sqrt{t^{2}-1}\right|+C

\displaystyle =-\log\left|(2-x)+\sqrt{(2-x)^{2}-1}\right|+C

\displaystyle \textbf{Question 10: }~\int \frac{x^4+1}{x^2+1}\,dx.
\displaystyle \text{Answer:}

\displaystyle I=\int \frac{x^{4}+1}{x^{2}+1}\,dx

\displaystyle =\int \frac{x^{4}-1+1+1}{x^{2}+1}\,dx

\displaystyle =\int \left[\frac{x^{4}-1}{x^{2}+1}+\frac{2}{x^{2}+1}\right]dx

\displaystyle =\int \left[\frac{(x^{2}-1)(x^{2}+1)}{x^{2}+1}+\frac{2}{x^{2}+1}\right]dx

\displaystyle =\int \left[(x^{2}-1)+\frac{2}{x^{2}+1}\right]dx

\displaystyle =\frac{x^{3}}{3}-x+2\tan^{-1}(x)+C

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