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Maths (commerce) Solve the Following Question.(2 Marks)

Integration (p-1) · Maharashtra Board STD 12 Commerce / Arts (MSBSHSE - English Medium). 25 questions.

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Solve the Following Question.(2 Marks)

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25 questions · timed · auto-graded

Question 12 Marks
Evalute : $\int \sqrt{x^2-8 x+7} d x$
Answer
$
\begin{aligned}
& \int \sqrt{x^2-8 x+7} d x \\
= & \int \sqrt{\left(x^2-8 x+16\right)-9} d x \\
= & \int \sqrt{(x-4)^2-(3)^2} d x \\
= & \frac{(x-4)}{2} \sqrt{(x-4)^2-(3)^2}- \\
& \frac{(3)^2}{2} \log \left|(x-4)+\sqrt{(x-4)^2-(3)^2}\right|+c \\
= & \frac{(x-4)}{2} \sqrt{x^2-8 x+7}- \\
& \frac{9}{2} \log \left|(x-4)+\sqrt{x^2-8 x+7}\right|+c .
\end{aligned}
$
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Question 22 Marks
Evalute : $\int \sqrt{x^2+2 x+5} d x$
Answer
$
\begin{aligned}
& \int \sqrt{x^2+2 x+5} d x \\
= & \int \sqrt{\left(x^2+2 x+1\right)+4} d x \\
= & \int \sqrt{(x+1)^2+(2)^2} d x \\
= & \frac{(x+1)}{2} \sqrt{(x+1)^2+(2)^2}+
\end{aligned}
$
$
\frac{(2)^2}{2} \log \left|(x+1)+\sqrt{(x+1)^2+(2)^2}\right|+c
$
$
=\frac{(x+1)}{2} \sqrt{x^2+2 x+5}+
$
$
2 \log \left|(x+1)+\sqrt{x^2+2 x+5}\right|+c .
$
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Question 32 Marks
Evalute : $\int e^{\sqrt{x}} d x$
Answer
$
\text { Let } I=\int e^{\sqrt{x}} d x
$
Put $\sqrt{x}=t \quad \therefore x=t^2$
$\therefore d x=2 t d t$
$
\begin{aligned}
\therefore I & =\int e^t \cdot 2 t d t=2 \int t e^t d t \\
& =2\left[t \int e^t d t-\int\left\{\frac{d}{d t}(t) \int e^t d t\right\}\right] d t \\
& =2\left[t \cdot e^t-\int 1 \cdot e^t d t\right] \\
& =2\left[t \cdot e^t-e^t\right]+c \\
& =2(t-1) e^t+c \\
& =2(\sqrt{x}-1) e^{\sqrt{x}}+c .
\end{aligned}
$
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Question 42 Marks
Evalute : $\int x e^{2 x} d x$
Answer
$
\begin{aligned}
\int x e^{2 x} d x & =x \int e^{2 x} d x-\int\left[\frac{d}{d x}(x) \int e^{2 x} d x\right] d x \\
& =x \cdot \frac{e^{2 x}}{2}-\int 1 \cdot \frac{e^{2 x}}{2} d x \\
& =\frac{1}{2} x e^{2 x}=\frac{1}{2} \int e^{2 x} d x \\
& =\frac{1}{2} x e^{2 x}-\frac{1}{2} \cdot \frac{e^{2 x}}{2}+c \\
& =e^{2 x}\left(\frac{x}{2}-\frac{1}{4}\right)+c \\
& =\left(\frac{2 x-1}{4}\right) e^{2 x}+c .
\end{aligned}
$
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Question 52 Marks
Evalute : $\int e^x \frac{1+x}{(2+x)^2} d x$
Answer
$
\begin{aligned}
& \text { Let } I=\int e^x \frac{1+x}{(2+x)^2} d x \\
& =\int e^x\left[\frac{(2+x)-1}{(2+x)^2}\right] d x \\
& =\int e^x\left[\frac{1}{2+x}-\frac{1}{(2+x)^2}\right] d x
\end{aligned}
$
Let $f(x)=\frac{1}{2+x}$
$
\begin{aligned}
& \therefore f^{\prime}(x)=\frac{-1}{(2+x)^2} \\
& \therefore I=\int e^x[f(x)+f \prime(x)] d x \\
& =e^x \cdot f(x)+c \\
& =e^x \cdot \frac{1}{2+x}+c \\
& \therefore I=\frac{e^x}{2+x}+c
\end{aligned}
$
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Question 62 Marks
Evalute : $\int(\log x)^2 d x$
Answer
$
\begin{aligned}
& \int(\log x)^2 d x=\int(\log x)^2 \cdot 1 d x \\
= & (\log x)^2 \int 1 d x-\int\left[\frac{d}{d x}(\log x)^2 \cdot \int 1 d x\right] d x \\
= & (\log x)^2 \cdot x-\int\left[2 \log x \cdot \frac{d}{d x}(\log x) \times x\right] d x \\
= & x(\log x)^2-\int 2 \log x \times \frac{1}{x} \times x d x \\
= & x(\log x)^2-2 \int(\log x) \cdot 1 d x \\
= & x(\log x)^2-2\left\{(\log x) \int 1 d x-\left[\frac{d}{d x}(\log x) \int 1 d x\right] d x\right\} \\
= & x(\log x)^2-2\left\{(\log x) \cdot x-\int \frac{1}{x} \times x d x\right\} \\
= & x(\log x)^2-2 x \log x+2 \int 1 d x \\
= & x(\log x)^2-2 x \log x+2 x+c .
\end{aligned}
$
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Question 72 Marks
Evalute : $\int \frac{e^x}{4 e^{2 x}-1} d x$
Answer
Let $I=\int \frac{e^x}{4 e^{2 x}-1} d x$
Put $e^x=t \quad \therefore e^x d x=d t$
$
\begin{aligned}
\therefore I & =\int \frac{1}{4 t^2-1} d t=\frac{1}{4} \int \frac{1}{ t ^2-\frac{1}{4}} d t \\
& =\frac{1}{4} \int \frac{1}{t^2-\left(\frac{1}{2}\right)^2} d t
\end{aligned}
$
$
\begin{aligned}
& =\frac{1}{4} \times \frac{1}{2 \times \frac{1}{2}} \log \left|\frac{t-\frac{1}{2}}{t+\frac{1}{2}}\right|+c \\
& =\frac{1}{4} \log \left|\frac{2 t-1}{2 t+1}\right|+c \\
& =\frac{1}{4} \log \left|\frac{2 e ^x-1}{2 e^x+1}\right|+c .
\end{aligned}
$
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Question 82 Marks
Evalute : $\int \frac{d x}{25 x-x(\log x)^2}$
Answer
$
\text { Let } \begin{aligned}
I & =\int \frac{d x}{25 x-x(\log x)^2} \\
& =\int \frac{1}{25-(\log x)^2} \cdot \frac{1}{x} d x
\end{aligned}
$
Put $\log x=t \quad \therefore \frac{1}{x} d x=d t$
$
\begin{aligned}
\therefore I & =\int \frac{1}{25-t^2} d t \\
& =\frac{1}{2 \times 5} \log \left|\frac{5+t}{5-t}\right|+c \\
& =\frac{1}{10} \log \left|\frac{5+\log x}{5-\log x}\right|+c .
\end{aligned}
$
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Question 102 Marks
Evalute : $\int \frac{e^x}{\sqrt{e^{2 x}+4 e^x+13}} d x$
Answer
Let $I=\int \frac{e^x}{\sqrt{e^{2 x}+4 e^x+13}} d x$
Put $e^x d x \quad \therefore e^x d x=d t$
$
\begin{aligned}
\therefore I & =\int \frac{1}{\sqrt{t^2+4 t+13}} d t \\
& =\int \frac{1}{\sqrt{\left(t^2+4 t+4\right)+9}} d t \\
& =\int \frac{1}{\sqrt{(t+2)^2+(3)^2}} d t \\
& =\log \left|(t+2)+\sqrt{(t+2)^2+(3)^2}\right|+c \\
& =\log \mid(t+2)+\sqrt{t^2+4 t+13 \mid}+c \\
& =\log \left|\left(e^x+2\right)+\sqrt{e^{2 x}+4 e^x+13}\right|+c .
\end{aligned}
$
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Question 112 Marks
Evalute : $\int \frac{1}{\sqrt{x}+x} d x$
Answer
$
\begin{aligned}
& \text { Let } I=\int \frac{1}{\sqrt{x}+x} d x \\
& =\int \frac{1}{\sqrt{x}(1+\sqrt{x})} d x
\end{aligned}
$
Put $1+\sqrt{ x }= t$
$
\begin{aligned}
& \therefore \frac{1}{2 \sqrt{ x }} dx = dt \\
& \therefore \frac{1}{\sqrt{ x }} dx =2 dt \\
& \therefore I =\int \frac{2 \cdot dt }{ t } \\
& =2 \int \frac{1}{ t } dt \\
& =2 \log | t |+ c \\
& \therefore I =2 \log |1+\sqrt{ x }|+ c
\end{aligned}
$
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Question 122 Marks
Evalute : $\int \frac{1}{2 x+3 x \log x} d x$
Answer
$
\text { Let } \begin{aligned}
I & =\int \frac{1}{2 x+3 x \log x} d x \\
& =\int \frac{1}{(2+3 \log x)} \cdot \frac{1}{x} d x
\end{aligned}
$
Put $2+3 \log x=t \quad \therefore \frac{3}{x} d x=d t$
$\therefore \frac{1}{x} d x=\frac{d t}{3}$
$
\begin{aligned}
\therefore I & =\int \frac{1}{t} \cdot \frac{d t}{3}=\frac{1}{3} \int \frac{1}{t} d t \\
& =\frac{1}{3} \log |t|+c \\
& =\frac{1}{3} \log |2+3 \log x|+c .
\end{aligned}
$
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Question 132 Marks
Evalute : Find the primitive of $\frac{1}{1+e^x}$
Answer
Let I be the primitive of $\frac{1}{1+e^x}$
$
\text { Then } \begin{aligned}
& I=\int \frac{1}{1+e^x} d x \\
&=\int \frac{\left(\frac{1}{e^x}\right)}{\left(\frac{1+e^x}{e^x}\right)} d x=\int \frac{e^{-x}}{e^{-x}+1} d x \\
&=-\int \frac{-e^{-x}}{e^{-x}+1} d x \\
&=-\log \left|e^{-x}+1\right|+c \\
& \quad \ldots\left[\because \frac{d}{d x}\left(e^{-x}+1\right)=-e^{-x}\right. \text { and } \\
&\left.\int \frac{f^{\prime}(x)}{f(x)} d x=\log |f(x)|+c\right]
\end{aligned}
$
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Question 142 Marks
Evalute : $\int\left[\frac{1}{\log x}-\frac{1}{(\log x)^2}\right] d x$
Answer
Let $I=\int\left[\frac{1}{\log x}-\frac{1}{(\log x)^2}\right] d x$
Put $\log x=t \quad \therefore x=e^t$
$\therefore d x=e^t d t$
$
\therefore I=\int\left(\frac{1}{t}-\frac{1}{t^2}\right) e^t d t
$
Let $f(t)=\frac{1}{t}$. Then $f^{\prime}(t)=-\frac{1}{t^2}$
$
\begin{aligned}
\therefore I & =\int e^t\left[f(t)+f^{\prime}(t)\right] d t \\
& =e^t \cdot f(t)+c=e^t \times \frac{1}{t}+c \\
& =\frac{x}{\log x}+c
\end{aligned}
$
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Question 152 Marks
Evalute : $\int e^x \frac{x-1}{(x+1)^3} d x$
Answer
$
\text { Let } \begin{aligned}
I & =\int e^x \cdot \frac{x-1}{(x+1)^3} d x \\
& =\int e^x\left[\frac{(x+1)-2}{(x+1)^3}\right] d x \\
& =\int e^x\left[\frac{1}{(x+1)^2}-\frac{2}{(x+1)^3}\right] d x
\end{aligned}
$
Put $f(x)=\frac{1}{(x+1)^2} $
Then $f^{\prime}(x)=\frac{d}{d x}(x+1)^{-2}=-2(x+1)^{-3} \cdot \frac{d}{d x}(x+1)$
$
\begin{gathered}
\quad=\frac{-2}{(x+1)^3} \times(1+0)=\frac{-2}{(x+1)^3} \\
\therefore I=\int e^x\left[f(x)+f^{\prime}(x)\right] d x \\
=e^x \cdot f(x)+c=e^x \cdot \frac{1}{(x+1)^2}+c
\end{gathered}
$
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Question 162 Marks
Evalute : $\int e^x \frac{x}{(x+1)^2} d x$
Answer
$
\begin{aligned}
LetI & =\int e^x \cdot \frac{x}{(x+1)^2} d x \\
& =\int e^x\left[\frac{(x+1)-1}{(x+1)^2}\right] d x \\
& =\int e^x\left[\frac{1}{x+1}-\frac{1}{(x+1)^2}\right] d x
\end{aligned}
$
Put $f(x)=\frac{1}{x+1}$
Then $f^{\prime}(x)=\frac{d}{d x}(x+1)^{-1}=-1(x+1)^{-2} \cdot \frac{d}{d x}(x+1)$
$
\begin{aligned}
& =\frac{-1}{(x+1)^2} \times(1+0)=\frac{-1}{(x+1)^2} \\
\therefore I=\int e^x[ & \left.f(x)+f^{\prime}(x)\right] d x \\
& =e^x \cdot f(x)+c=e^x \cdot \frac{1}{x+1}+c .
\end{aligned}
$
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Question 172 Marks
Evalute : $\int x^3 e^{x^2} d x$
Answer
Let $I=\int x^3 e^{x^2} d x=\int x^2 e^{x^2} \cdot x d x$
Put $x^2=t \quad \therefore 2 x d x=d t$
$
\begin{aligned}
& \therefore x d x=\frac{d t}{2}\\
& \therefore I=\int t e^t \cdot \frac{d t}{2}=\frac{1}{2} \int t e^t d t \\
& =\frac{1}{2}\left[t \int e^t d t-\int\left\{\frac{d}{d t}(t) \int e^t d t\right\} d t\right] \\
& =\frac{1}{2}\left[t e^t-\int 1 \cdot e^t d t\right] \\
& =\frac{1}{2}\left[t e^t-e^t\right]+c \\
& =\frac{1}{2}(t-1) e^t+c \\
& =\frac{1}{2}\left(x^2-1\right) e^{x^2}+c . \\
\end{aligned}
$
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Question 182 Marks
Evalute : $\int \frac{1}{a^2-b^2 x^2} d x$
Answer
$
\begin{aligned}
\int \frac{1}{a^2-b^2 x^2} d x=\frac{1}{b^2} \int \frac{1}{\frac{a^2}{b^2}-x^2} d x\\
=\frac{1}{b^2} \int \frac{1}{\left(\frac{a}{b}\right)^2-x^2} d x \\
=\frac{1}{b^2} \times \frac{1}{2\left(\frac{a}{b}\right)} \log \left|\frac{\frac{a}{b}+x}{\frac{a}{b}-x}\right|+c \\
=\frac{1}{2 a b} \log \left|\frac{a+b x}{a-b x}\right|+c .
\end{aligned}
$
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Question 192 Marks
Evalute : $\int \frac{1}{x^2+4 x-5} d x$
Answer
$
\begin{aligned}
& \int \frac{1}{x^2+4 x-5} d x \\
= & \int \frac{1}{\left(x^2+4 x+4\right)-4-5} d x \\
= & \int \frac{1}{(x+2)^2-(3)^2} d x \\
= & \frac{1}{2 \times 3} \log \left|\frac{x+2-3}{x+2+3}\right|+c \\
= & \frac{1}{6} \log \left|\frac{x-1}{x+5}\right|+c .
\end{aligned}
$
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Question 202 Marks
Evalute : $\int \frac{1}{4 x^2-1} d x$
Answer
$
\begin{aligned}
& \quad \int \frac{1}{4 x^2-1} d x=\frac{1}{4} \int \frac{1}{x^2-(1 / 4)} d x \\
& =\frac{1}{4} \int \frac{1}{x^2-\left(\frac{1}{2}\right)^2} d x
\end{aligned}
$
$
\begin{aligned}
& =\frac{1}{4} \times \frac{1}{2\left(\frac{1}{2}\right)} \log \left|\frac{x-\frac{1}{2}}{x+\frac{1}{2}}\right|+c \\
& =\frac{1}{4} \log \left|\frac{2 x-1}{2 x+1}\right|+c .
\end{aligned}
$
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Question 212 Marks
Evalute : $\int \frac{1}{\sqrt{x}+x} d x$
Answer
$
\text { Let } \begin{aligned}
I & =\int \frac{1}{\sqrt{x}+x} d x=\int \frac{1}{\sqrt{x}(1+\sqrt{x})} d x \\
& =\int \frac{1}{1+\sqrt{x}} \cdot \frac{1}{\sqrt{x}} d x
\end{aligned}
$
Put $1+\sqrt{x}=t \quad \therefore \frac{1}{2 \sqrt{x}} d x=d t$
$
\begin{aligned}
\therefore & \frac{1}{\sqrt{x}} d x=2 d t \\
\therefore I & =\int \frac{1}{t} \cdot 2 d t=2 \int \frac{1}{t} d t \\
& =2 \log |t|+c=2 \log |1+\sqrt{x}|+c .
\end{aligned}
$
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Question 222 Marks
Evalute : $\int \frac{1+x}{x+e^{-x}} d x$
Answer
$
\text { Let } \begin{aligned}
I & =\int \frac{1+x}{x+e^{-x}} d x \\
& =\int \frac{(1+x) e^x}{\left(x+e^{-x}\right) e^x} d x \\
& =\int \frac{(1+x) e^x}{x e^x+1} d x
\end{aligned}
$
Put $x e^x+1=t$
$
\begin{aligned}
& \therefore\left(x e^x+e^x \times 1\right) d x=d t \\
& \therefore(1+x) e^x d x=d t \\
& \therefore I=\int \frac{1}{t} d t=\log |t|+c \\
& \quad=\log \left|x e^x+1\right|+c .
\end{aligned}
$
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Question 232 Marks
Evalute : $\int \frac{x^3}{\sqrt{1+x^4}} d x$
Answer
Let $I=\int \frac{x^3}{\sqrt{1+x^4}} d x$
Put $1+x^4=t \quad \therefore 4 x^3 d x=d t$
$
\begin{aligned}
& \therefore x^3 d x=\frac{d t}{4} \\
& \begin{aligned}
\therefore I & =\int \frac{1}{\sqrt{t}} \cdot \frac{d t}{4}=\frac{1}{4} \int t^{-\frac{1}{2}} d t \\
& =\frac{1}{4} \cdot \frac{t^{\frac{1}{2}}}{\left(\frac{1}{2}\right)}+c \\
& =\frac{1}{2} \sqrt{1+x^4}+c .
\end{aligned}
\end{aligned}
$
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Question 242 Marks
Evalute : $\int x \sqrt{1+x^2} d x$
Answer
$
\text { Let } I=\int x \sqrt{1+x^2} d x=\int \sqrt{1+x^2} \cdot x d x
$
Put $1+x^2=t$
$
\begin{aligned}
& \therefore 2 x d x=d t \quad \therefore x d x=\frac{d t}{2} \\
& \begin{aligned}
\therefore I & =\int \sqrt{t} \frac{d t}{2}=\frac{1}{2} \int t^{\frac{1}{2}} d t \\
& =\frac{1}{2} \cdot \frac{t^{\frac{3}{2}}}{(3 / 2)}+c \\
& =\frac{1}{3}\left(1+x^2\right)^{\frac{3}{2}}+c .
\end{aligned}
\end{aligned}
$
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Question 252 Marks
Evaluate $\int \frac{-2}{\sqrt{5 x-4}-\sqrt{5 x-2}} d x$
Answer
$
\begin{aligned}
& \int \frac{-2}{\sqrt{5 x-4}-\sqrt{5 x-2}} d x \\
= & \int \frac{-2}{\sqrt{5 x-4}-\sqrt{5 x-2}} \times \frac{\sqrt{5 x-4}+\sqrt{5 x-2}}{\sqrt{5 x-4}+\sqrt{5 x-2}} d x \\
= & \int \frac{-2(\sqrt{5 x-4}+\sqrt{5 x-2})}{(5 x-4)-(5 x-2)} d x \\
= & \int(\sqrt{5 x-4}+\sqrt{5 x-2}) d x \\
= & \int(\sqrt{5 x-4})^{\frac{1}{2}} d x+\int(\sqrt{5 x-2})^{\frac{1}{2}} d x \\
= & \frac{(5 x-4)^{\frac{3}{2}}}{\left(\frac{3}{2}\right)} \times \frac{1}{5}+\frac{(5 x-2)^{\frac{3}{2}}}{\left(\frac{3}{2}\right)} \times \frac{1}{5}+c \\
= & \frac{2}{15}\left[(5 x-4)^{\frac{3}{2}}+(5 x-2)^{\frac{3}{2}}\right]+c .
\end{aligned}
$
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