\displaystyle \text{Evaluate the following integrals:}

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

\displaystyle \int x^2\sqrt{x+2}\,dx

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

\displaystyle \Rightarrow x=t-2

\displaystyle \Rightarrow dx=dt

\displaystyle \int x^2\sqrt{x+2}\,dx=\int (t-2)^2\sqrt{t}\,dt

\displaystyle =\int (t^2-4t+4)t^{\tfrac12}\,dt

\displaystyle =\int \left(t^{\tfrac52}-4t^{\tfrac32}+4t^{\tfrac12}\right)\,dt

\displaystyle =\left[\frac{t^{\tfrac72}}{\tfrac72}-4\frac{t^{\tfrac52}}{\tfrac52}+4\frac{t^{\tfrac32}}{\tfrac32}\right]+C

\displaystyle =\frac{2}{7}t^{\tfrac72}-\frac{8}{5}t^{\tfrac52}+\frac{8}{3}t^{\tfrac32}+C

\displaystyle =\frac{2}{7}(x+2)^{\tfrac72}-\frac{8}{5}(x+2)^{\tfrac52}+\frac{8}{3}(x+2)^{\tfrac32}+C

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

\displaystyle \int \frac{x^2}{\sqrt{x-1}}\,dx

\displaystyle \text{Let } x-1=t^2

\displaystyle \Rightarrow x=t^2+1

\displaystyle \Rightarrow dx=2t\,dt

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

\displaystyle =2\int (t^2+1)^2\,dt

\displaystyle =2\int (t^4+2t^2+1)\,dt

\displaystyle =2\left(\frac{t^5}{5}+\frac{2t^3}{3}+t\right)+C

\displaystyle =\frac{2t^5}{5}+\frac{4t^3}{3}+2t+C

\displaystyle =\frac{2}{15}(3t^5+10t^3+15t)+C

\displaystyle =\frac{2}{15}\,t\,(3t^4+10t^2+15)+C

\displaystyle =\frac{2}{15}\sqrt{x-1}\,[3(x-1)^2+10(x-1)+15]+C

\displaystyle =\frac{2}{15}\sqrt{x-1}\,(3x^2+4x+8)+C

\displaystyle \textbf{Question 3: }~\int \frac{x^2}{\sqrt{3x+4}}\,dx.
\displaystyle \text{Answer:}

\displaystyle \int \frac{x^2}{\sqrt{3x+4}}\,dx

\displaystyle \text{Let } 3x+4=t

\displaystyle \Rightarrow x=\frac{t-4}{3}

\displaystyle \Rightarrow dx=\frac{dt}{3}

\displaystyle \int \frac{x^2}{\sqrt{3x+4}}\,dx =\frac{1}{3}\int \frac{\left(\frac{t-4}{3}\right)^2}{\sqrt{t}}\,dt

\displaystyle =\frac{1}{27}\int \frac{(t-4)^2}{\sqrt{t}}\,dt

\displaystyle =\frac{1}{27}\int \left(t^{\tfrac32}-8t^{\tfrac12}+16t^{-\tfrac12}\right)\,dt

\displaystyle =\frac{1}{27}\left(\frac{t^{\tfrac52}}{\tfrac52}-8\frac{t^{\tfrac32}}{\tfrac32}+16\frac{t^{\tfrac12}}{\tfrac12}\right)+C

\displaystyle =\frac{2}{135}t^{\tfrac52}-\frac{16}{81}t^{\tfrac32}+\frac{32}{27}t^{\tfrac12}+C

\displaystyle =\frac{2}{135}(3x+4)^{\tfrac52}-\frac{16}{81}(3x+4)^{\tfrac32}+\frac{32}{27}(3x+4)^{\tfrac12}+C

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

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

\displaystyle \text{Let } x-1=t

\displaystyle \Rightarrow x=t+1

\displaystyle \Rightarrow dx=dt

\displaystyle \int \frac{2x-1}{(x-1)^2}\,dx =\int \frac{2(t+1)-1}{t^2}\,dt

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

\displaystyle =\int \left(\frac{2}{t}+\frac{1}{t^2}\right)\,dt

\displaystyle =2\log|t|-\frac{1}{t}+C

\displaystyle =2\log(x-1)-\frac{1}{x-1}+C

\displaystyle \textbf{Question 5: }~\int (2x^2+3)\sqrt{x+2}\,dx.
\displaystyle \text{Answer:}

\displaystyle \int (2x^2+3)\sqrt{x+2}\,dx

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

\displaystyle \Rightarrow x=t-2

\displaystyle \Rightarrow dx=dt

\displaystyle \int (2x^2+3)\sqrt{x+2}\,dx =\int \left[2(t-2)^2+3\right]\sqrt{t}\,dt

\displaystyle =\int (2t^2-8t+11)\,t^{\tfrac12}\,dt

\displaystyle =\int \left(2t^{\tfrac52}-8t^{\tfrac32}+11t^{\tfrac12}\right)\,dt

\displaystyle =\frac{4}{7}t^{\tfrac72}-\frac{16}{5}t^{\tfrac52}+\frac{22}{3}t^{\tfrac32}+C

\displaystyle =\frac{4}{7}(x+2)^{\tfrac72}-\frac{16}{5}(x+2)^{\tfrac52}+\frac{22}{3}(x+2)^{\tfrac32}+C

\displaystyle \textbf{Question 6: }~\int \frac{x^2+3x+1}{(x+1)^2}\,dx.
\displaystyle \text{Answer:}

\displaystyle \int \frac{x^2+3x+1}{(x+1)^2}\,dx

\displaystyle \text{Let } x+1=t

\displaystyle \Rightarrow x=t-1

\displaystyle \Rightarrow dx=dt

\displaystyle \int \frac{x^2+3x+1}{(x+1)^2}\,dx =\int \frac{(t-1)^2+3(t-1)+1}{t^2}\,dt

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

\displaystyle =\int \left(1+\frac{1}{t}-\frac{1}{t^2}\right)\,dt

\displaystyle =t+\log|t|+\frac{1}{t}+C

\displaystyle =x+1+\log|x+1|+\frac{1}{x+1}+C

\displaystyle =x+\log|x+1|+\frac{1}{x+1}+C

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

\displaystyle \text{Let } I=\int \frac{x^2}{\sqrt{x-1}}\,dx

\displaystyle \text{Substituting } x-1=t \text{ and } dx=dt

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

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

\displaystyle =\int \left(t^{\tfrac32}+2t^{\tfrac12}+t^{-\tfrac12}\right)\,dt

\displaystyle =\frac{2}{5}t^{\tfrac52}+2t^{\tfrac32}+2t^{\tfrac12}+C

\displaystyle =\frac{2}{15}\left(3t^{\tfrac52}+15t^{\tfrac32}+15t^{\tfrac12}\right)+C

\displaystyle =\frac{2}{15}\,t^{\tfrac12}\left(3t^2+15+10t\right)+C

\displaystyle =\frac{2}{15}\sqrt{x-1}\,[3(x-1)^2+15+10(x-1)]+C

\displaystyle =\frac{2}{15}\sqrt{x-1}\,(3x^2+4x+8)+C

\displaystyle \textbf{Question 8: }~\int x(1-x)^{23}\,dx.
\displaystyle \text{Answer:}

\displaystyle \int x(1-x)^{23}\,dx

\displaystyle \text{Let } 1-x=t

\displaystyle \Rightarrow x=1-t

\displaystyle \Rightarrow dx=-dt

\displaystyle \int x(1-x)^{23}\,dx=-\int (1-t)t^{23}\,dt

\displaystyle =-\int (t^{23}-t^{24})\,dt

\displaystyle =\int (t^{24}-t^{23})\,dt

\displaystyle =\frac{t^{25}}{25}-\frac{t^{24}}{24}+C

\displaystyle =\frac{1}{600}\left(24t^{25}-25t^{24}\right)+C

\displaystyle =\frac{t^{24}}{600}(24t-25)+C

\displaystyle =\frac{(1-x)^{24}}{600}\left(24(1-x)-25\right)+C

\displaystyle =-\frac{(1-x)^{24}}{600}(1+24x)+C

\displaystyle \textbf{Question 9: }~\int \frac{1}{\sqrt{x}+\sqrt[4]{x}}\,dx.
\displaystyle \text{Answer:}

\displaystyle \text{Let } I=\int \frac{1}{\sqrt{x}+\sqrt[4]{x}}\,dx

\displaystyle \text{Let } x=t^4

\displaystyle \Rightarrow dx=4t^3\,dt

\displaystyle I=\int \frac{4t^3}{\sqrt{t^4}+\sqrt[4]{t^4}}\,dt

\displaystyle =\int \frac{4t^3}{t^2+t}\,dt

\displaystyle =4\int \frac{t^2}{t+1}\,dt

\displaystyle =4\int \frac{(t-1)(t+1)+1}{t+1}\,dt

\displaystyle =4\int \left(t-1+\frac{1}{t+1}\right)\,dt

\displaystyle =4\left(\frac{t^2}{2}-t+\log(t+1)\right)+C

\displaystyle =2t^2-4t+4\log(t+1)+C

\displaystyle =2\sqrt{x}-4\sqrt[4]{x}+4\log(\sqrt[4]{x}+1)+C

\displaystyle \textbf{Question 10: }~\int \frac{1}{x^{1/3}\left(x^{1/3}-1\right)}\,dx.
\displaystyle \text{Answer:}

\displaystyle \text{Let } I=\int \frac{1}{x^{\tfrac13}(x^{\tfrac13}-1)}\,dx

\displaystyle \text{Let } x=t^3

\displaystyle \Rightarrow dx=3t^2\,dt

\displaystyle I=\int \frac{3t^2}{(t^3)^{\tfrac13}\left((t^3)^{\tfrac13}-1\right)}\,dt

\displaystyle =\int \frac{3t^2}{t(t-1)}\,dt

\displaystyle =3\int \frac{t}{t-1}\,dt

\displaystyle =3\int \frac{(t-1)+1}{t-1}\,dt

\displaystyle =3\int \left(1+\frac{1}{t-1}\right)\,dt

\displaystyle =3\left(t+\log|t-1|\right)+C

\displaystyle =3x^{\tfrac13}+3\log\!\left|x^{\tfrac13}-1\right|+C

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