\displaystyle \textbf{Question 1: }\int \frac{x+1}{\sqrt{2x+3}}\,dx.
\displaystyle \text{Answer:}

\displaystyle \int \left(\frac{x+1}{\sqrt{2x+3}}\right)\,dx

\displaystyle = \frac{1}{2}\int \left(\frac{2x+2}{\sqrt{2x+3}}\right)\,dx

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

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

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

\displaystyle = \frac{1}{2}\left[\int (2x+3)^{\tfrac{1}{2}}\,dx-\int (2x+3)^{-\tfrac{1}{2}}\,dx\right]

\displaystyle = \frac{1}{2}\left[\frac{(2x+3)^{\tfrac{3}{2}}}{3}- (2x+3)^{\tfrac{1}{2}}\right]+C

\displaystyle = \frac{1}{6}(2x+3)^{\tfrac{3}{2}}-\frac{1}{2}(2x+3)^{\tfrac{1}{2}}+C

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

\displaystyle \text{Let } I=\int x\sqrt{x+2}\,dx

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

\displaystyle \text{Then, } x=t-2

\displaystyle \text{Differentiating both sides, } dx=dt

\displaystyle \text{Now, the integral becomes}

\displaystyle I=\int (t-2)\sqrt{t}\,dt

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

\displaystyle =\left[\frac{t^{\tfrac{5}{2}}}{\tfrac{5}{2}}-2\cdot\frac{t^{\tfrac{3}{2}}}{\tfrac{3}{2}}\right]+C

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

\displaystyle =\frac{2}{5}(x+2)^{\tfrac{5}{2}}-\frac{4}{3}(x+2)^{\tfrac{3}{2}}+C

\displaystyle \textbf{Question 3: }\int \frac{x-1}{\sqrt{x+4}}\,dx.
\displaystyle \text{Answer:}

\displaystyle \text{Let } I=\int \left(\frac{x-1}{\sqrt{x+4}}\right)\,dx

\displaystyle \text{Putting } x+4=t

\displaystyle \text{Then, } x=t-4

\displaystyle \text{Differentiating both sides, } dx=dt

\displaystyle \text{Now, the integral becomes}

\displaystyle I=\int \left(\frac{t-5}{\sqrt{t}}\right)\,dt

\displaystyle =\int \left(t^{\tfrac{1}{2}}-5t^{-\tfrac{1}{2}}\right)\,dt

\displaystyle =\left[\frac{t^{\tfrac{3}{2}}}{\tfrac{3}{2}}-5\cdot\frac{t^{\tfrac{1}{2}}}{\tfrac{1}{2}}\right]+C

\displaystyle =\frac{2}{3}t^{\tfrac{3}{2}}-10t^{\tfrac{1}{2}}+C

\displaystyle =\frac{2}{3}(x+4)^{\tfrac{3}{2}}-10(x+4)^{\tfrac{1}{2}}+C

\displaystyle \textbf{Question 4: }\int (x+2)\sqrt{3x+5}\,dx.
\displaystyle \text{Answer:}

\displaystyle \text{Let } I=\int (x+2)\sqrt{3x+5}\,dx

\displaystyle \text{Putting } 3x+5=t

\displaystyle \Rightarrow x=\frac{t-5}{3}

\displaystyle \Rightarrow 3\,dx=dt

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

\displaystyle \text{Therefore,}

\displaystyle I=\int \left(\frac{t-5}{3}+2\right)\sqrt{t}\cdot\frac{dt}{3}

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

\displaystyle =\frac{1}{9}\int \left(t^{\tfrac{3}{2}}+t^{\tfrac{1}{2}}\right)\,dt

\displaystyle =\frac{1}{9}\left[\frac{t^{\tfrac{5}{2}}}{\tfrac{5}{2}}+\frac{t^{\tfrac{3}{2}}}{\tfrac{3}{2}}\right]+C

\displaystyle =\frac{1}{9}\left[\frac{2}{5}t^{\tfrac{5}{2}}+\frac{2}{3}t^{\tfrac{3}{2}}\right]+C

\displaystyle =\frac{2}{45}t^{\tfrac{5}{2}}+\frac{2}{27}t^{\tfrac{3}{2}}+C

\displaystyle =\frac{2}{45}(3x+5)^{\tfrac{5}{2}}+\frac{2}{27}(3x+5)^{\tfrac{3}{2}}+C

\displaystyle \textbf{Question 5: }\int \frac{2x+1}{\sqrt{3x+2}}\,dx.
\displaystyle \text{Answer:}

\displaystyle \int \left(\frac{2x+1}{\sqrt{3x+2}}\right)\,dx

\displaystyle =\frac{1}{3}\int \left(\frac{6x+3}{\sqrt{3x+2}}\right)\,dx

\displaystyle =\frac{1}{3}\int \left(\frac{6x+4-1}{\sqrt{3x+2}}\right)\,dx

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

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

\displaystyle =\frac{1}{3}\left[2\int (3x+2)^{\tfrac{1}{2}}\,dx-\int (3x+2)^{-\tfrac{1}{2}}\,dx\right]

\displaystyle =\frac{1}{3}\left[2\cdot\frac{(3x+2)^{\tfrac{3}{2}}}{3\cdot\tfrac{3}{2}}-\frac{(3x+2)^{\tfrac{1}{2}}}{3\cdot\tfrac{1}{2}}\right]+C

\displaystyle =\frac{1}{3}\left[\frac{4}{9}(3x+2)^{\tfrac{3}{2}}-\frac{2}{3}(3x+2)^{\tfrac{1}{2}}\right]+C

\displaystyle =\frac{4}{27}(3x+2)^{\tfrac{3}{2}}-\frac{2}{9}(3x+2)^{\tfrac{1}{2}}+C

\displaystyle =\frac{2}{27}(6x+1)\sqrt{3x+2}+C

\displaystyle \textbf{Question 6: }\int \frac{3x+5}{\sqrt{7x+9}}\,dx.
\displaystyle \text{Answer:}

\displaystyle \text{Let } I=\int \left(\frac{3x+5}{\sqrt{7x+9}}\right)\,dx

\displaystyle \text{Putting } 7x+9=t

\displaystyle \Rightarrow x=\frac{t-9}{7}

\displaystyle \Rightarrow 7\,dx=dt

\displaystyle \Rightarrow dx=\frac{dt}{7}

\displaystyle \text{Therefore,}

\displaystyle I=\int \left(\frac{3\left(\frac{t-9}{7}\right)+5}{\sqrt{t}}\right)\cdot\frac{dt}{7}

\displaystyle =\frac{1}{7}\int \left(\frac{3t-27+35}{7\sqrt{t}}\right)\,dt

\displaystyle =\frac{1}{49}\int \left(\frac{3t+8}{\sqrt{t}}\right)\,dt

\displaystyle =\frac{1}{49}\int \left(3t^{\tfrac{1}{2}}+8t^{-\tfrac{1}{2}}\right)\,dt

\displaystyle =\frac{1}{49}\left[3\cdot\frac{t^{\tfrac{3}{2}}}{\tfrac{3}{2}}+8\cdot\frac{t^{\tfrac{1}{2}}}{\tfrac{1}{2}}\right]+C

\displaystyle =\frac{1}{49}\left[2t^{\tfrac{3}{2}}+16t^{\tfrac{1}{2}}\right]+C

\displaystyle =\frac{2}{49}t^{\tfrac{3}{2}}+\frac{16}{49}t^{\tfrac{1}{2}}+C

\displaystyle =\frac{2}{49}(7x+9)^{\tfrac{3}{2}}+\frac{16}{49}(7x+9)^{\tfrac{1}{2}}+C

\displaystyle =\frac{2}{49}(7x+9)^{\tfrac{1}{2}}(7x+17)+C

\displaystyle \textbf{Question 7: }\int \frac{x}{\sqrt{x+4}}\,dx.
\displaystyle \text{Answer:}

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

\displaystyle =\int \left(\frac{x+4-4}{\sqrt{x+4}}\right)\,dx

\displaystyle =\int \left(\sqrt{x+4}-\frac{4}{\sqrt{x+4}}\right)\,dx

\displaystyle =\int (x+4)^{\tfrac{1}{2}}\,dx-4\int (x+4)^{-\tfrac{1}{2}}\,dx

\displaystyle =\frac{(x+4)^{\tfrac{3}{2}}}{\tfrac{3}{2}}-4\cdot\frac{(x+4)^{\tfrac{1}{2}}}{\tfrac{1}{2}}+C

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

\displaystyle =(x+4)^{\tfrac{1}{2}}\left[\frac{2}{3}(x+4)-8\right]+C

\displaystyle =(x+4)^{\tfrac{1}{2}}\left(\frac{2x-16}{3}\right)+C

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

\displaystyle \textbf{Question 8: }\int \frac{2-3x}{\sqrt{1+3x}}\,dx.
\displaystyle \text{Answer:}

\displaystyle \text{Let } I=\int \left(\frac{2-3x}{\sqrt{1+3x}}\right)\,dx

\displaystyle \text{Putting } 1+3x=t

\displaystyle \Rightarrow 3x=t-1

\displaystyle \text{and } 3\,dx=dt

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

\displaystyle \text{Therefore,}

\displaystyle I=\int \left(\frac{2-(t-1)}{\sqrt{t}}\right)\cdot\frac{dt}{3}

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

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

\displaystyle =\frac{1}{3}\left[3\cdot\frac{t^{\tfrac{1}{2}}}{\tfrac{1}{2}}-\frac{t^{\tfrac{3}{2}}}{\tfrac{3}{2}}\right]+C

\displaystyle =2t^{\tfrac{1}{2}}-\frac{2}{9}t^{\tfrac{3}{2}}+C

\displaystyle =2\sqrt{t}-\frac{2}{9}t\sqrt{t}+C

\displaystyle =2\sqrt{1+3x}-\frac{2}{9}(1+3x)\sqrt{1+3x}+C

\displaystyle =\frac{2}{9}(8-3x)\sqrt{1+3x}+C

\displaystyle \textbf{Question 9: }\int (5x+3)\sqrt{2x-1}\,dx.
\displaystyle \text{Answer:}

\displaystyle \text{Let } I=\int (5x+3)\sqrt{2x-1}\,dx

\displaystyle \text{Putting } 2x-1=t

\displaystyle \Rightarrow 2x=t+1

\displaystyle \Rightarrow x=\frac{t+1}{2}

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

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

\displaystyle \text{Therefore,}

\displaystyle I=\int \left[5\left(\frac{t+1}{2}\right)+3\right]\sqrt{t}\cdot\frac{dt}{2}

\displaystyle =\int \left(\frac{5t+11}{2}\right)\sqrt{t}\cdot\frac{dt}{2}

\displaystyle =\frac{1}{4}\int (5t+11)t^{\tfrac{1}{2}}\,dt

\displaystyle =\frac{1}{4}\int \left(5t^{\tfrac{3}{2}}+11t^{\tfrac{1}{2}}\right)\,dt

\displaystyle =\frac{1}{4}\left[\frac{5t^{\tfrac{5}{2}}}{\tfrac{5}{2}}+\frac{11t^{\tfrac{3}{2}}}{\tfrac{3}{2}}\right]+C

\displaystyle =\frac{1}{2}t^{\tfrac{5}{2}}+\frac{11}{6}t^{\tfrac{3}{2}}+C

\displaystyle =\frac{t^{\tfrac{3}{2}}}{2}\left(t+\frac{11}{3}\right)+C

\displaystyle =\frac{t^{\tfrac{3}{2}}}{2}\left(\frac{3t+11}{3}\right)+C

\displaystyle =\frac{(2x-1)^{\tfrac{3}{2}}}{2}\left(\frac{3(2x-1)+11}{3}\right)+C

\displaystyle =\frac{(2x-1)^{\tfrac{3}{2}}}{2}\left(\frac{6x+8}{3}\right)+C

\displaystyle =\frac{(2x-1)^{\tfrac{3}{2}}(3x+4)}{3}+C

\displaystyle \textbf{Question 10: }\int \frac{x}{\sqrt{x+a}-\sqrt{x+b}}\,dx.
\displaystyle \text{Answer:}

\displaystyle \int \frac{x}{\sqrt{x+a}-\sqrt{x+b}}\,dx

\displaystyle =\int \frac{x}{\sqrt{x+a}-\sqrt{x+b}}\cdot\frac{\sqrt{x+a}+\sqrt{x+b}}{\sqrt{x+a}+\sqrt{x+b}}\,dx

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

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

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

\displaystyle =\frac{1}{a-b}\left[\int x\sqrt{x+a}\,dx+\int x\sqrt{x+b}\,dx\right]

\displaystyle =\frac{1}{a-b}\left[\int (x+a-a)\sqrt{x+a}\,dx+\int (x+b-b)\sqrt{x+b}\,dx\right]

\displaystyle =\frac{1}{a-b}\Bigg[\int (x+a)\sqrt{x+a}\,dx-a\int \sqrt{x+a}\,dx

\displaystyle \qquad\qquad\qquad\qquad+\int (x+b)\sqrt{x+b}\,dx-b\int \sqrt{x+b}\,dx\Bigg]

\displaystyle =\frac{1}{a-b}\Bigg[\int (x+a)^{\tfrac{3}{2}}\,dx-a\int (x+a)^{\tfrac{1}{2}}\,dx

\displaystyle \qquad\qquad\qquad\qquad+\int (x+b)^{\tfrac{3}{2}}\,dx-b\int (x+b)^{\tfrac{1}{2}}\,dx\Bigg]

\displaystyle =\frac{1}{a-b}\Bigg[\frac{(x+a)^{\tfrac{5}{2}}}{\tfrac{5}{2}}-a\frac{(x+a)^{\tfrac{3}{2}}}{\tfrac{3}{2}}

\displaystyle \qquad\qquad\qquad\qquad+\frac{(x+b)^{\tfrac{5}{2}}}{\tfrac{5}{2}}-b\frac{(x+b)^{\tfrac{3}{2}}}{\tfrac{3}{2}}\Bigg]+C

\displaystyle =\frac{1}{a-b}\left[\frac{2}{5}(x+a)^{\tfrac{5}{2}}-\frac{2a}{3}(x+a)^{\tfrac{3}{2}}+\frac{2}{5}(x+b)^{\tfrac{5}{2}}-\frac{2b}{3}(x+b)^{\tfrac{3}{2}}\right]+C

Discover more from ICSE / ISC / CBSE Mathematics Portal for K12 Students

Subscribe to get the latest posts sent to your email.