-8747-x-3x-42ex -160-dx-a- exx-4-C-b- exx-3-C-c- 1x-42-C-d- exx-42-C

(a) exx+4+CLet #160;I= #8747;x+3x+42ex #160;dx #8658; #8747;x+4 1x+42 #160;ex #160;dx #8658; #8747;1x+4 1x+42 #160;ex #160;dxAs, #160;we ...

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Byju's Answer

Standard VII

Mathematics

Properties of Integers Using Literals

#8747;x+3x+42...

Question

$\int \frac{x + 3}{{(x + 4)}^{2}} e^{x} d x =$

(a) $\frac{e^{x}}{x + 4} + C$

(b) $\frac{e^{x}}{x + 3} + C$

(c) $\frac{1}{{(x + 4)}^{2}} + C$

(d) $\frac{e^{x}}{{(x + 4)}^{2}} + C$

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Solution

(a) $\frac{e^{x}}{x + 4} + C$

$Let I = \int \frac{(x + 3)}{{(x + 4)}^{2}} e^{x} d x \Rightarrow \int [\frac{x + 4 - 1}{{(x + 4)}^{2}}] e^{x} d x \Rightarrow \int [\frac{1}{(x + 4)} - \frac{1}{{(x + 4)}^{2}}] e^{x} d x As, we know that \int e^{x} \{f (x) + f' (x)\} d x = e^{x} f (x) + C ∴ I = \frac{e^{x}}{x + 4} + C$

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∫x+3x+42ex dx= (a) exx+4+C (b) exx+3+C (c) 1x+42+C (d) exx+42+C

(a) exx+4+CLet I=∫x+3x+42ex dx⇒∫x+4-1x+42 ex dx⇒∫1x+4-1x+42 ex dxAs, we know that ∫exfx+f'x dx=exfx+C∴ I=exx+4+C

$\int \frac{x + 3}{{(x + 4)}^{2}} e^{x} d x =$

(a) $\frac{e^{x}}{x + 4} + C$

(b) $\frac{e^{x}}{x + 3} + C$

(c) $\frac{1}{{(x + 4)}^{2}} + C$

(d) $\frac{e^{x}}{{(x + 4)}^{2}} + C$