Question

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Solve this using - Indefinite Integration Type 3:

∫ √1 + 3x - x² dx.


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Answer 1

Answer :

Now,

$\int \sqrt{1 + 3x - {x}^{2} } dx \\ \\ = \int \sqrt{(1 + \frac{9}{4}) - ( {x}^{2} - 3x + \frac{9}{4} )} \: \: dx \\ \\ = \int \sqrt{ {( \frac{ \sqrt{13} }{2}) }^{2} - {(x - \frac{3}{2} )}^{2} } \: \: dx \\ \\ = \frac{(x - \frac{3}{2} ) \sqrt{ {( \frac{ \sqrt{13} }{2}) ^{2} - {(x - \frac{3}{2} )} }^{2} } }{2} \\ + \frac{ {( \frac{ \sqrt{13} }{2} )}^{2} }{2} \: sin ^{ - 1} \frac{(x - \frac{3}{2} )}{ (\frac{ \sqrt{13} }{2}) } + c, \\ \\ where \: \: c \: \: is \: \: integral \: \: constant \\ \\ = \frac{(2x - 3 ) \sqrt{1 + 3x - {x}^{2} } }{4} \\ + \frac{13}{8} \: {sin}^{ - 1} ( \frac{2x - 3}{ \sqrt{13} } ) + c \\ \\ RULE : \\ \\ \int \sqrt{ {a}^{2} - {x}^{2} } \: \: dx \\ \\ = \frac{x \sqrt{ {a}^{2} - {x}^{2} } }{2} + \frac{ {a}^{2} }{2} \: \: {sin}^{ - 1} ( \frac{x}{a} ) + c, \\ \\ where \: \: c \: \: is\: \: integral \: \: constant$

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Answer 2

Step-by-step explanation:

Let z = 2x3 + 3x

dz = (6x2 +3)dx = 3( 2x2 +1)dx

    = 1/3 ∫ sin2z dz = 1/3 ∫ 1–cos2z/2 dz

    = 1/6 ∫ (1–cos2z)dz = 1/6 [z – sin2z / z] + c

    = 1/6 [2x2 + 3x –sin(4x3 + 6x)/2] = + c