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Can someone solve these integrals?

+3 votes
a) ∫x^2-3x+2/√x dx

b) ∫e^2 dx

c) ∫(x+1)^3/2x dx

d) ∫dx/4x^2+25
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asked Dec 28, 2012 in CALCULUS by chrisgirl Apprentice

5 Answers

+3 votes

a) ∫ (x^2-3x+2/√x) dx

= ∫x^2 dx - ∫3xdx + ∫2/√x dx

= ∫x^2 dx -  3∫xdx +2 ∫1/√x dx

The integral of ∫x^n dx = (x^n+1)/(n+1) + c

= (x^2+1) / (2+1) - 3 (x^1+1)/(1+1) + 2 (x^ -1/2+1)/( -1/2+1 )+ c

= (x^3) / 3 - 3 ( x^2 )/2 + 2 ( x^ 1/2)/ (1/2) + c

= (x^3) / 3 - 3 ( x^2 ) /2 + 4 (x^1/2) + c

answered Dec 28, 2012 by ashokavf Scholar
+3 votes

b) ∫e^2 dx

= ∫e^2 dx

= e^2∫ 1 dx

= e^2 ( x + c)

= x e^2 + e^2 c

answered Dec 28, 2012 by ashokavf Scholar

∫e2 dx

Integral of e2 is e2x.

= e2x + C

+3 votes

d) ∫dx/4x^2+25

= ∫(1/4x^2+25)dx

= ∫(1/(2x)^2+(5)^2)dx

Using integral formula ∫(1/(x)^2+(a)^2)dx = (1/a)tan−1 (x / a) + c

= tan−1 (2x / 5) d/dx (2x / 5)  + c

=(2/5) tan−1 (2x / 5) + c

answered Dec 28, 2012 by ashokavf Scholar

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Take 25 as common.

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+3 votes

a) ∫ (x^2-3x+2/√x) dx

= ∫x^2 dx - ∫3xdx + ∫2/√x dx

= ∫x^2 dx -  3∫xdx +2 ∫1/√x dx

The integral of ∫x^n dx = (x^n+1)/(n+1) + c

= (x^2+1) / (2+1) - 3 (x^1+1)/(1+1) + 2 (x^ -1/2+1)/( -1/2+1 )+ c

= (x^3) / 3 - 3 ( x^2 )/2 + 2 ( x^ 1/2)/ (1/2) + c

= (x^3) / 3 - 3 ( x^2 ) /2 + 4 (x^1/2) + c

=2x^3 - 9x^2  +  24 x^1/2 + 6c

answered Dec 29, 2012 by krish Pupil
0 votes

c)

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Do long division.

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answered Jun 23, 2014 by joly Scholar

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