mathematics-class-12-unit-07-chapter-12-indefinite-integral-l-5-7-1b0gked-qrs

<h3 id="success-stylefont-size25px-indefinite-integral-l-5-success-1"><Success style="font-size:25px"> Indefinite Integral L-5 </Success></h3>
<h3 id="primary-stylefont-size24px-problem-1-primary"><primary style="font-size:24px"> Problem-1 </primary></h3>
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<p>$I=\int \frac{2 x-1}{\sqrt{4 x-x^2}} d x $</p>
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<p>Solution:</p>
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<p>$2 x-1=A \frac{d}{d x}\left(4 x-x^2\right)+B $</p>
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<p>$ 2 x-1=A(4-2 x)+B $</p>
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<p>$ 2=-2 A \Rightarrow A=-1,$</p>
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<p>$4 A+B=-1 \Rightarrow B=3 $</p>
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<p>$ I=\int \frac{-1(4-2 x)+3}{\sqrt{4 x-x^2}} d x $</p>
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<p>$ =-\int \frac{4-2 x}{\sqrt{4 x-x^2}} d x+3 \int \frac{d x}{\sqrt{4 x-x^2}}$</p>
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<p>[$ 4 x-x^2=-\left(x^2-4 x\right)=4-\left(x^2-4 x+4\right) =4-(x-2)^2 $ ]</p>
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<footer style = "font-size: 18px"> $\rightarrow$ Indefinite integral lecture-12 $\rightarrow$ <span style = "color:blue"> Problem-1 </span> $\rightarrow$ Solution $\rightarrow$ Problem-2 </footer>

<h3 id="success-stylefont-size25px-indefinite-integral-l-5-success-2"><Success style="font-size:25px"> Indefinite Integral L-5 </Success></h3>
<h3 id="primary-stylefont-size24px-solution-primary"><primary style="font-size:24px"> Solution </primary></h3>
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<li>[$ 4 x-x^2=t \Rightarrow(4-2 x) d x=d t $]</li>
<li>$ I=-\int \frac{d t}{\sqrt{t}}+3 \int \frac{d x}{\sqrt{4-(x-2)^2}}$</li>
<li>[$x-2=u \Rightarrow  d x=d u $]</li>
<li>$ =-\frac{t^{1 / 2}}{1 / 2}+3 \int \frac{d u}{\sqrt{4-u^2}}$</li>
<li>$ =-2 \sqrt{t}+3 \sin ^{-1} \frac{u}{2}+C $</li>
<li>$ =-2 \sqrt{4 x-x^2}+3 \sin ^{-1}\left(\frac{x-2}{2}\right)+C $</li>
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<footer style = "font-size: 18px">Indefinite integral lecture-12 $\rightarrow$ Problem-1 $\rightarrow$ <span style = "color:blue"> Solution </span> $\rightarrow$ Problem-2 $\rightarrow$ Method of partial fractions </footer>

### <Success style="font-size:25px"> Indefinite Integral L-5 </Success>
### <primary style="font-size:24px"> Problem-2 </primary>
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- $ I=\int \frac{x^2 d x}{\sqrt{x^6+a^6}} \quad a>0$

- Solution:

- $=\int \frac{x^2 d x}{\sqrt{\left(x^3\right)^2+a^6}}$

- [$ x^3=t \Rightarrow 3 x^2 d x=d t] $

- $=\int \frac{1}{3} \frac{d t}{\sqrt{t^2+\left(a^3\right)^2}} $

- $=\frac{1}{3}.\log \left|t+\sqrt{\left.t^2+a^3\right)^2}\right|+c $

- $ =\frac{1}{3} \log \left|x^3+\sqrt{x^6+a^6}\right|+c $

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<footer style = "font-size: 18px">Problem-1 $\rightarrow$ Solution $\rightarrow$ <span style = "color:blue"> Problem-2 </span> $\rightarrow$ Method of partial fractions $\rightarrow$ Method of partial fraction forms </footer>

### <Success style="font-size:25px"> Indefinite Integral L-5 </Success>
### <primary style="font-size:24px"> Method of partial fraction forms </primary>
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- Froms

- 1. $\frac{p x+q}{(x-a)(x-b)} \quad a \neq b$

- 2. $\frac{p x+q}{(x-a)^2}$

- 3. $\frac{p x^2+q x+r}{(x-a)(x-b)(x-c)},$   
- $  a \neq b, b \neq c \\ a \neq c $

- 4. $\frac{p x^2+q x+r}{\left.(x-a)^2(x-b\right)}$

- 5. $\frac{p x^2+a x+r}{(x-a_1)\left(a x^2+b x+c\right)}$

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- Partial fractions

- 1 $ \frac{A}{x-a}+\frac{B}{x-b} $

- 2 $ \frac{A}{x-a}+\frac{B}{(x-a)^2} $

- 3 $ \frac{A}{x-a}+\frac{B}{x-b}+\frac{c}{x-c} $

- 4 $ \frac{A}{(x-a)}+\frac{B}{(x-a)^2}+\frac{C}{x-b} $

- 5 $ \frac{A}{x-a_1}+\frac{B x+C}{a x^2+b x+c}$

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<footer style = "font-size: 18px">Problem-2 $\rightarrow$ Method of partial fractions $\rightarrow$ <span style = "color:blue"> Method of partial fraction forms </span> $\rightarrow$ Problem-3 $\rightarrow$ Problem-4 </footer>

### <Success style="font-size:25px"> Indefinite Integral L-5 </Success>
### <primary style="font-size:24px"> Problem-3 </primary>
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- $   \int \frac{d x}{x^2-a^2}=\frac{1}{2 a} \log |\frac{x-a}{x+a}|+C$

- Solution:
- $  \frac{1}{(x-a)(x+a)}=\frac{A}{(x-a)}+\frac{B}{(x+a)}$

- $  \Rightarrow  1=A(x+a)+B(x-a)$

- $  A+B=0, \quad a(A-B)=1 $  
- $  A=-B, \quad-2 a B=1$  
- $  \Rightarrow B=-\frac{1}{2 a} \cdot A=\frac{1}{2 a} $   
- $ \frac{1}{(x-a)(x+a)}  =\frac{1}{2 a}\left[\frac{1}{x-a}-\frac{1}{x+a}\right] $   
- $  =\frac{1}{2 a}[\log |x-a|-\log |x+a|]+c$
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<footer style = "font-size: 18px">Method of partial fractions $\rightarrow$ Method of partial fraction forms $\rightarrow$ <span style = "color:blue"> Problem-3 </span> $\rightarrow$ Problem-4 $\rightarrow$ Problem-5 </footer>

### <Success style="font-size:25px"> Indefinite Integral L-5 </Success>
### <primary style="font-size:24px"> Problem-4 </primary>
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- $\int \frac{d x}{x\left(x^4+1\right)}$

- Solution
- $= \int \frac{x^3 d x}{x^4\left(x^4+1\right)}$

- [$ x^4=t \Rightarrow  4 x^3 d x=d t \Rightarrow x^3 d x=\frac{d t}{4}$

- $=  \frac{1}{4} \int \frac{d t}{t(t+1)}$

- $\\{\frac{1}{t(t+1)}=\frac{A}{t}+\frac{B}{t+1} \Rightarrow 1=A(t+1)+B t \Rightarrow A=1, B=-1 \\}$

- $ =\frac{1}{4} \int\left(\frac{1}{t}-\frac{1}{t+1}\right) d t$   
- $=\frac{1}{4}\left[{[\log |t|-\log |t+1|]} +C\right. $
- $=\frac{1}{4}[{\log |\frac{x^4}{x^4 + 1}|}] +C. $

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<footer style = "font-size: 18px">Method of partial fraction forms $\rightarrow$ Problem-3 $\rightarrow$ <span style = "color:blue"> Problem-4 </span> $\rightarrow$ Problem-5 $\rightarrow$ Solution </footer>

### <Success style="font-size:25px"> Indefinite Integral L-5 </Success>
### <primary style="font-size:24px"> Solution </primary>
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- $ =\int 1 d x-\\{\int \frac{-2 d x}{x^2+3}+\int \frac{6 d x}{x^2+4}\\}$

- $  =x+2 \int \frac{1}{x^2+3} d x-6 \int \frac{1}{x^2+4} d x $

- $ =x + 2.\frac{1}{\sqrt{3}} \tan ^{-1} \frac{x}{\sqrt{3}} - 6.\frac{1}{2}.\tan ^{-1} \frac{x}{2}+C $
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<footer style = "font-size: 18px">Problem-5 $\rightarrow$ Solution $\rightarrow$ <span style = "color:blue"> Solution </span> $\rightarrow$ Problem-6 $\rightarrow$ Integration by parts </footer>

### <Success style="font-size:25px"> Indefinite Integral L-5 </Success>
### <primary style="font-size:24px"> Problem-6 </primary>
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- $I=\int \frac{x+1}{(x-2)\left(x^2+1\right)} d x$

- $ \frac{x+1}{(x-2)\left(x^2+1\right)}=\frac{A}{x-2}+\frac{B x+C}{x^2+1} $

- $ \Rightarrow x+1=A\left(x^2+1\right)+(B x+C)(x-2) $

- $ A+B=0,-2 B+C=1, A-2 C=1 $

- $ A=\frac{3}{5}, B=-\frac{3}{5}, C=-1 / 5 $

- $ I=\int\left[\frac{3}{5} \cdot \frac{1}{x-2}+\frac{-\frac{3}{5} x-\frac{1}{5}}{x^2+1}\right] d x $

- $ =\frac{3}{5} \int \frac{d x}{x-2}-\frac{1}{5} \int \frac{3 x d x}{x^2+1}-\frac{1}{5} \int \frac{1}{x^2+1} d x$

- $=\frac{3}{5} \log |x-2|-\frac{3}{5} \times \frac{1}{2} \log \left|x^2+1\right|-\frac{1}{5} \tan ^{-1} x+C\$

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<footer style = "font-size: 18px">Solution $\rightarrow$ Solution $\rightarrow$ <span style = "color:blue"> Problem-6 </span> $\rightarrow$ Integration by parts $\rightarrow$ Problem-7 </footer>

### <Success style="font-size:25px"> Indefinite Integral L-5 </Success>
### <primary style="font-size:24px"> Integration by parts </primary>
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- $   \frac{d}{d x}(u v)=u \frac{d v}{d x}+\frac{d u}{d x} v $

- $  u v=\int u \frac{d v}{d x} d x+\int \frac{d u}{d x} v d x $

- $   \int u \frac{d v}{d x} d x=u v-\int \frac{d u}{d x} v d x $

- $   u=f(x) ; \quad \frac{d v}{d x}=g(x) $

- $   \int f(x) g(x) d x=f(x) \int g(x) d x-\int [f^{\prime}(x).\int g(x)] d x d x$

- f(x) = 1st function   
- g(x) = second function

- Int. = first function X integral of second fn - Integral {diff. of first fn X Integral of second fn}

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<footer style = "font-size: 18px">Solution $\rightarrow$ Problem-6 $\rightarrow$ <span style = "color:blue"> Integration by parts </span> $\rightarrow$ Problem-7 $\rightarrow$ Thankyou </footer>

### <Success style="font-size:25px"> Indefinite Integral L-5 </Success>
### <primary style="font-size:24px"> Problem-7 </primary>
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- $ I  =\int x e^x d x $  
- $x$ - 1st function    
- $e^x$ - 2nd function    
- $I =x \cdot e^x-\int 1 \cdot e^x d x $
- $I =x e^x-e^x+c$

- Remark:-

- $ \int x e^x d x $

- = $ x \cdot\left(e^x+c\right)-\int 1 \cdot\left(e^x+c\right) d x $

- = $ x e^x+c x-e^x-c  x+c_1 $

- = $ x e^x-e^x+c_1$  
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<footer style = "font-size: 18px">Problem-6 $\rightarrow$ Integration by parts $\rightarrow$ <span style = "color:blue"> Problem-7  </span> $\rightarrow$ Thankyou $\rightarrow$  </footer>