solve this
Question: $\int \frac{1}{x \sqrt{1+x^{2}}} d x$ Solution: Let $x=\sin ^{\frac{2}{3}} t$ Differentiate both side with respect to $t$ $\frac{d x}{d t}=\frac{2}{3} \sin ^{-\frac{1}{3}} t \cos t \Rightarrow d x=\frac{2}{3} \sin ^{-\frac{1}{3}} t \cos t d t$ $\frac{d x}{d t}=\frac{2}{3} \sin ^{-\frac{1}{3}} t \cos t \Rightarrow d x=\frac{2}{3} \sin ^{-\frac{1}{3}} t \cos t d t$ $y=\int \frac{1}{\sin ^{\frac{2}{3}} t \sqrt{1+\sin ^{2} t}} \frac{2}{3} \sin ^{-\frac{1}{3}} t \cos t d t$ $y=\frac{2}{3} \...
Read More →Mark the correct alternative in each of the following:
Question: Mark the correct alternative in each of the following: If $\int \frac{1}{(x+2)\left(x^{2}+1\right)} d x a \log \mid 1+x^{2}+b \tan ^{-1}$ $\mathrm{x}+\frac{1}{5} \log |\mathrm{x}+2|+\mathrm{C}$, then A. $\mathrm{a}=-\frac{1}{10}, \mathrm{~b}=-\frac{2}{5}$ B. $a=\frac{1}{10}, b=-\frac{2}{5}$ C. $a=-\frac{1}{10}, b=\frac{2}{5}$ D. $a=\frac{1}{10}, b=\frac{2}{5}$ Solution: $U=\int \frac{1}{(x+2)\left(x^{2}+1\right)} d x$ $U=\int \frac{A}{x+2} d x+\int \frac{B x+c}{x^{2}+1} d x$ $\frac{1}{...
Read More →Mark the correct alternative in each of the following:
Question: Mark the correct alternative in each of the following: If $\int \frac{1}{(x+2)\left(x^{2}+1\right)} d x a \log \mid 1+x^{2}+b \tan ^{-1}$ $\mathrm{x}+\frac{1}{5} \log |\mathrm{x}+2|+\mathrm{C}$, then A. $\mathrm{a}=-\frac{1}{10}, \mathrm{~b}=-\frac{2}{5}$ B. $a=\frac{1}{10}, b=-\frac{2}{5}$ C. $a=-\frac{1}{10}, b=\frac{2}{5}$ D. $a=\frac{1}{10}, b=\frac{2}{5}$ Solution: $U=\int \frac{1}{(x+2)\left(x^{2}+1\right)} d x$ $U=\int \frac{A}{x+2} d x+\int \frac{B x+c}{x^{2}+1} d x$ $\frac{1}{...
Read More →Mark the correct alternative in each of the following:
Question: Mark the correct alternative in each of the following: $\int \frac{x^{3}}{x+1} d x$ A. $x+\frac{x^{2}}{2}+\frac{x^{3}}{3}-\log |1-x|+C$ B. $x+\frac{x^{2}}{2}-\frac{x^{3}}{3}-\log |1-x|+C$ C. $x-\frac{x^{2}}{2}-\frac{x^{3}}{3}-\log |1+x|+C$ D. $x-\frac{x^{2}}{2}+\frac{x^{3}}{3}-\log |1+x|+C$ Solution: $=\int \frac{x^{3}+1}{x+1} d x-\int \frac{1}{x+1} d x$ $=\int \frac{(x+1)\left(x^{2}-x+1\right)}{x+1} d x-\int \frac{1}{x+1} d x$ $=\int\left(x^{2}-x+1\right) d x-\int \frac{1}{x+1} d x$ $...
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Question: $f(x)=\left\{\begin{array}{ll}\frac{2^{x+2}-16}{4^{x}-16}, \text { if } x \neq 2 \\ k, \text { if } x=2\end{array}\right.$ at $x=2$ Solution: The given function f(x) can be rewritten as, $f(x)=\frac{2^{x+2}-16}{4^{x}-16}=\frac{2^{2} \cdot 2^{x}-16}{\left(2^{x}\right)^{2}-(4)^{2}}=\frac{4\left(2^{x}-4\right)}{\left(2^{x}-4\right)\left(2^{x}+4\right)}$ $f(x)=\frac{4}{2^{x}+4}$ $\lim _{x \rightarrow 2^{-}} f(x)=\lim _{h \rightarrow 0} \frac{4}{2^{2-h}+4}=\frac{4}{2^{2}+4}=\frac{4}{4+4}=\f...
Read More →Mark the correct alternative in each of the following:
Question: Mark the correct alternative in each of the following: $\int \frac{x^{3}}{x+1} d x$ A. $x+\frac{x^{2}}{2}+\frac{x^{3}}{3}-\log |1-x|+C$ B. $x+\frac{x^{2}}{2}-\frac{x^{3}}{3}-\log |1-x|+C$ C. $x-\frac{x^{2}}{2}-\frac{x^{3}}{3}-\log |1+x|+C$ D. $x-\frac{x^{2}}{2}+\frac{x^{3}}{3}-\log |1+x|+C$ Solution: $=\int \frac{x^{3}+1}{x+1} d x-\int \frac{1}{x+1} d x$ $=\int \frac{(x+1)\left(x^{2}-x+1\right)}{x+1} d x-\int \frac{1}{x+1} d x$ $=\int\left(x^{2}-x+1\right) d x-\int \frac{1}{x+1} d x$ $...
Read More →Mark the correct alternative in each of the following:
Question: Mark the correct alternative in each of the following: $\int \frac{x^{3}}{x+1} d x$ A. $x+\frac{x^{2}}{2}+\frac{x^{3}}{3}-\log |1-x|+C$ B. $x+\frac{x^{2}}{2}-\frac{x^{3}}{3}-\log |1-x|+C$ C. $x-\frac{x^{2}}{2}-\frac{x^{3}}{3}-\log |1+x|+C$ D. $x-\frac{x^{2}}{2}+\frac{x^{3}}{3}-\log |1+x|+C$ Solution: $=\int \frac{x^{3}+1}{x+1} d x-\int \frac{1}{x+1} d x$ $=\int \frac{(x+1)\left(x^{2}-x+1\right)}{x+1} d x-\int \frac{1}{x+1} d x$ $=\int\left(x^{2}-x+1\right) d x-\int \frac{1}{x+1} d x$ $...
Read More →Mark the correct alternative in each of the following:
Question: Mark the correct alternative in each of the following: $\int \frac{x^{3}}{\sqrt{1+x^{2}}} d x=a\left(1+x^{2}\right)^{3 / 2}+b \sqrt{1+x^{2}}+C$, then A. $a=\frac{1}{3}, b=1$ B. $a=-\frac{1}{3}, b=1$ C. $a=-\frac{1}{3}, b=-1$ D. $a=\frac{1}{3}, b=-1$ Solution: let $\left(\sqrt{1+x^{2}}\right)=\mathrm{t}$ $\frac{x}{\sqrt{1+x^{2}}} d x=d t$ $I=\int \frac{x^{3}}{\sqrt{1+x^{2}}} d x=\int x^{2} d t=\int\left(t^{2}-1\right) d t$ $I=\frac{t^{3}}{3}-t\left[\operatorname{put}(t)=\sqrt{1+x^{2}}\r...
Read More →Mark the correct alternative in each of the following:
Question: Mark the correct alternative in each of the following: $\int \frac{x^{3}}{\sqrt{1+x^{2}}} d x=a\left(1+x^{2}\right)^{3 / 2}+b \sqrt{1+x^{2}}+C$, then A. $a=\frac{1}{3}, b=1$ B. $a=-\frac{1}{3}, b=1$ C. $a=-\frac{1}{3}, b=-1$ D. $a=\frac{1}{3}, b=-1$ Solution: let $\left(\sqrt{1+x^{2}}\right)=\mathrm{t}$ $\frac{x}{\sqrt{1+x^{2}}} d x=d t$ $I=\int \frac{x^{3}}{\sqrt{1+x^{2}}} d x=\int x^{2} d t=\int\left(t^{2}-1\right) d t$ $I=\frac{t^{3}}{3}-t\left[\operatorname{put}(t)=\sqrt{1+x^{2}}\r...
Read More →Mark the correct alternative in each of the following:
Question: Mark the correct alternative in each of the following: $\int \frac{x^{3}}{\sqrt{1+x^{2}}} d x=a\left(1+x^{2}\right)^{3 / 2}+b \sqrt{1+x^{2}}+C$, then A. $a=\frac{1}{3}, b=1$ B. $a=-\frac{1}{3}, b=1$ C. $a=-\frac{1}{3}, b=-1$ D. $a=\frac{1}{3}, b=-1$ Solution: let $\left(\sqrt{1+x^{2}}\right)=\mathrm{t}$ $\frac{x}{\sqrt{1+x^{2}}} d x=d t$ $I=\int \frac{x^{3}}{\sqrt{1+x^{2}}} d x=\int x^{2} d t=\int\left(t^{2}-1\right) d t$ $I=\frac{t^{3}}{3}-t\left[\right.$ put $\left.(t)=\sqrt{1+x^{2}}...
Read More →Mark the correct alternative in each of the following:
Question: Mark the correct alternative in each of the following: $\int \frac{x^{3}}{\sqrt{1+x^{2}}} d x=a\left(1+x^{2}\right)^{3 / 2}+b \sqrt{1+x^{2}}+C$, then A. $a=\frac{1}{3}, b=1$ B. $a=-\frac{1}{3}, b=1$ C. $a=-\frac{1}{3}, b=-1$ D. $a=\frac{1}{3}, b=-1$ Solution: let $\left(\sqrt{1+x^{2}}\right)=\mathrm{t}$ $\frac{x}{\sqrt{1+x^{2}}} d x=d t$ $I=\int \frac{x^{3}}{\sqrt{1+x^{2}}} d x=\int x^{2} d t=\int\left(t^{2}-1\right) d t$ $I=\frac{t^{3}}{3}-t\left[\right.$ put $\left.(t)=\sqrt{1+x^{2}}...
Read More →Mark the correct alternative in each of the following:
Question: Mark the correct alternative in each of the following: $\int \frac{x^{9}}{\left(4 x^{2}+1\right)^{6}} d x$ is equal to A. $\frac{1}{5 x}\left(4+\frac{1}{x^{2}}\right)^{-5}+C$ B. $\frac{1}{5}\left(4+\frac{1}{\mathrm{x}^{2}}\right)^{-5}+\mathrm{C}$ C. $\frac{1}{10 x}\left(\frac{1}{x^{2}}+4\right)^{-5}+C$ D. $\frac{1}{10}\left(\frac{1}{x^{2}}+4\right)^{-5}+C$ Solution: $\mathrm{I}=\int \frac{x^{9}}{\left(4 x^{2}+1\right)^{6}} d x$ $I=\int \frac{x^{9}}{x^{12}\left(4+\frac{1}{x^{2}} 6\right...
Read More →Mark the correct alternative in each of the following:
Question: Mark the correct alternative in each of the following: $\int \frac{x^{9}}{\left(4 x^{2}+1\right)^{6}} d x$ is equal to A. $\frac{1}{5 \mathrm{x}}\left(4+\frac{1}{\mathrm{x}^{2}}\right)^{-5}+\mathrm{C}$ B. $\frac{1}{5}\left(4+\frac{1}{x^{2}}\right)^{-5}+C$ C. $\frac{1}{10 \mathrm{x}}\left(\frac{1}{\mathrm{x}^{2}}+4\right)^{-5}+\mathrm{C}$ D. $\frac{1}{10}\left(\frac{1}{x^{2}}+4\right)^{-5}+C$ Solution: $\mathrm{I}=\int \frac{x^{9}}{\left(4 x^{2}+1\right)^{6}} d x$ $I=\int \frac{x^{9}}{x...
Read More →Mark the correct alternative in each of the following:
Question: Mark the correct alternative in each of the following: $\int \frac{\cos 2 x-\cos 2 \theta}{\cos x-\cos \theta} d x$ is equal to A. $2(\sin x+x \cos \theta)+C$ B. $2(\sin x-x \cos \theta)+C$ C. $2(\sin x+2 x \cos \theta)+C$ D. $2(\sin x-2 x \cos \theta)+C$ Solution: $I=\int \frac{\left\{2(\cos x)^{2}-1\right)-\left[2(\cos \theta)^{2}-1\right]}{\cos x-\cos \theta} \mathrm{d}_{x}$ $I=2 \int \frac{(\cos x)^{2}-(\cos \theta)^{2}}{\cos x-\cos \theta} d x$ $I=2 \int \frac{(\cos x-\cos \theta)...
Read More →Mark the correct alternative in each of the following:
Question: Mark the correct alternative in each of the following: $\int \frac{\cos 2 x-\cos 2 \theta}{\cos x-\cos \theta} d x$ is equal to A. $2(\sin x+x \cos \theta)+c$ B. $2(\sin x-x \cos \theta)+C$ C. $2(\sin x+2 x \cos \theta)+C$ D. $2(\sin x-2 x \cos \theta)+C$ Solution: $I=\int \frac{\left\{2(\cos x)^{2}-1\right\}-\left\{2(\cos \theta)^{2}-1\right\}}{\cos x-\cos \theta} d x$ $I=2 \int \frac{(\cos x)^{2}-(\cos \theta)^{2}}{\cos x-\cos \theta} d x$ $I=2 \int \frac{(\cos x-\cos \theta)(\cos x+...
Read More →Mark the correct alternative in each of the following:
Question: Mark the correct alternative in each of the following: $\int \frac{\cos 2 x-1}{\cos 2 x+1} d x=$ A. $\tan x-x+C$ B. $x+\tan x+C$ C. $x-\tan x+C$ D. $-x-\cot x+C$ Solution: $I=\int \frac{1-2(\sin x)^{2}-1}{2(\cos x)^{2}-1+1}$ $I=-\int \frac{(\sin x)^{2}}{(\cos x)^{2}} d x$ $I=-\int(\tan x)^{2} d x$ $I=-\int\left(-1+(\sec x)^{2} d x\right.$ $=(x-\tan x)+c$...
Read More →Mark the correct alternative in each of the following:
Question: Mark the correct alternative in each of the following: $\int \frac{\cos 2 x-1}{\cos 2 x+1} d x=$ A. $\tan x-x+C$ B. $x+\tan x+C$ C. $x-\tan x+C$ D. $-x-\cot x+C$ Solution: $I=\int \frac{1-2(\sin x)^{2}-1}{2(\cos x)^{2}-1+1}$ $I=-\int \frac{(\sin x)^{2}}{(\cos x)^{2}} d x$ $I=-\int(\tan x)^{2} d x$ $I=-\int\left(-1+(\sec x)^{2} d x\right.$ $=(x-\tan x)+c$...
Read More →Mark the correct alternative in each of the following:
Question: Mark the correct alternative in each of the following: If $\int x \sin x d x=-x \cos x+\alpha$, then $\alpha$ is equal to A. $\sin x+C$ B. $\cos x+C$ C. $\mathrm{C}$ D. none of these Solution: using integration by parts $I=\int x \sin x d \square$ $\left.=x \int \sin x d x-\int \frac{d x}{d x}(x) \int \sin x\right)$ $I=x \cos x+\int \cos x d x$ $\left(\because \int \sin x=-\cos x\right)$ $=x \cos x+\sin x+c$...
Read More →Mark the correct alternative in each of the following:
Question: Mark the correct alternative in each of the following: If $\int x \sin x d x=-x \cos x+\alpha$, then $\alpha$ is equal to A. $\sin x+c$ B. $\cos x+C$ C. $\mathrm{C}$ D. none of these Solution: using integration by parts $I=\int x \sin x d$ $\left.=x \int \sin x d x-\int \frac{d x}{d x}(x) \int \sin x\right)$ $I=x \cos x+\int \cos x d x$ $\left(\because \int \sin x=-\cos x\right)$ $=x \cos x+\sin x+c$...
Read More →Mark the correct alternative in each of the following:
Question: Mark the correct alternative in each of the following: The value of $\int \frac{\sin x+\cos x}{\sqrt{1-\sin 2 x}} d x$ is equal to A. $\sqrt{\sin 2 x}+C$ B. $\sqrt{\cos 2 x}+C$ C. $\pm(\sin x-\cos x)+C$ D. $\pm \log (\sin x-\cos x)+C$ Solution: $I=\int \frac{\sin x+\cos x}{\sin x-\cos x} d x(\sqrt{1-\sin 2 x}=\pm\{\sin x-\cos x\})$ Let $\mathrm{t}=\sin \mathrm{x}-\cos \times\left(\frac{d t}{d x}=\sin x+\cos x\right)$ $I=\int \frac{d t}{t}$ $I=\pm \log (\sin x-\cos x)+c$...
Read More →Mark the correct alternative in each of the following:
Question: Mark the correct alternative in each of the following: The value of $\int \frac{\sin x+\cos x}{\sqrt{1-\sin 2 x}} d x$ is equal to A. $\sqrt{\sin 2 x}+C$ B. $\sqrt{\cos 2 x}+C$ C. $\pm(\sin x-\cos x)+C$ D. $\pm \log (\sin x-\cos x)+C$ Solution: $I=\int \frac{\sin x+\cos x}{\sin x-\cos x} d x(\sqrt{1-\sin 2 x}=\pm\{\sin x-\cos x\})$ Let $\mathrm{t}=\sin x-\cos x\left(\frac{d t}{d x}=\sin x+\cos x\right)$ $I=\int \frac{d t}{t}$ $I=\pm \log (\sin x-\cos x)+c$...
Read More →Mark the correct alternative in each of the following:
Question: Mark the correct alternative in each of the following: $\int e^{x}\left\{f(x)+f^{\prime}(x)\right\} d x=$ A. $e^{x} f(x)+C$ B. $e^{x}+f(x)+C$ c. $2 e^{x} f(x)+C$ D. $e^{x}-f(x)+C$ Solution: let $\mathrm{I}=\int e^{x}\left(f(x)+f^{\prime}(x)\right) \mathrm{d}_{x}$ Open the brackets, we get $I=\left\{\int e^{x} f(x) d x+\int e^{x} f^{\prime}(x) d x\right\}$ $=U+\int e^{x} f^{\prime}(x) d x$ $U=\int e^{x} f(x) d x$ To solve U using integration by parts $U=f(x) \int e^{x} d x-\int\left[f^{...
Read More →Mark the correct alternative in each of the following:
Question: Mark the correct alternative in each of the following: $\int e^{x}\left\{f(x)+f^{\prime}(x)\right\} d x=$ A. $e^{x} f(x)+C$ B. $e^{x}+f(x)+C$ c. $2 e^{x} f(x)+c$ D. $e^{x}-f(x)+C$ Solution: let $\mathrm{I}=\int e^{x}\left(f(x)+f^{\prime}(x)\right) \mathrm{d}_{x}$ Open the brackets, we get $I=\left\{\int e^{x} f(x) d x+\int e^{x} f^{\prime}(x) d x\right\}$ $=U+\int e^{x} f^{\prime}(x) d x$ $U=\int e^{x} f(x) d x$ To solve $U$ using integration by parts $U=f(x) \int e^{x} d x-\int\left[f...
Read More →Mark the correct alternative in each of the following:
Question: Mark the correct alternative in each of the following: $\int \sqrt{\frac{x}{1-x}} d x$ is equal to A. $\sin ^{-1} \sqrt{x}+C$ B. $\sin ^{-1}(\sqrt{x}-\sqrt{x(1-x)})+C$ C. $\sin ^{-1}\{\sqrt{x(1-x)}\}+C$ D. $\sin ^{-1} \sqrt{x}-\sqrt{x(1-x)}+C$ Solution: let $x=(\sin t)^{2} ;(d x=2 \sin t \cos t d t)$ $I=\int \sqrt{\frac{(\sin t)^{2}}{1-(\sin t)^{2}}} \times 2 \sin t \cos t d t$ $I=\int(\sin t)^{2} d t$ $I=\int(1-\cos 2 t) d t$ $I=\int 1 d t-\int \cos 2 t d t$ $I=t-\frac{\sin 2 t}{2}+c\...
Read More →Mark the correct alternative in each of the following:
Question: Mark the correct alternative in each of the following: $\int \sqrt{\frac{x}{1-x}} d x$ is equal to A. $\sin ^{-1} \sqrt{\mathrm{x}}+\mathrm{C}$ B. $\sin ^{-1}(\sqrt{x}-\sqrt{x(1-x)})+C$ C. $\sin ^{-1}\{\sqrt{x(1-x)}\}+C$ D. $\sin ^{-1} \sqrt{x}-\sqrt{x(1-x)}+C$ Solution: let $x=(\sin t)^{2} ;(d x=2 \sin t \cos t d t)$ $I=\int \sqrt{\frac{(\sin t)^{2}}{1-(\sin t)^{2}}} \times 2 \sin t \cos t d t$ $\mathrm{I}=\int(\sin \mathrm{t})^{2} \mathrm{dt}$ $\mathrm{I}=\int(1-\cos 2 \mathrm{t}) \m...
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