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Evaluate d∫x2+xx4-9dx - Mathematics

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Question

Evaluate `int (x^2 + x)/(x^4 - 9) "d"x`

Sum

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Solution

We have I = `int (x^2 + x)/(x^4 - 9) "d"x`

= `int x^2/(x^4 - 9) "d"x + (x"d"x)/(x^4 - 9)`

= I₁ + I₂

Now I₁ = int x^3/(x^4 - 9)`

Put t = x⁴ – 9

So that 4x³ dx = dt.

Therefore I₁ = `1/4 int "dt"/"t"`

= `1/4 log|"t"| + "C"_1`

= `1/4 log|x^4 - 9| + "C"_1`

Again, I₂ = `int (x"d"x)/(x^4 - 9)`

Put x² = u

So that 2x dx = du

Then I₂ = `1/2 int "du"/("u"^2 - (3)^2)`

= `1/(2 xx 6) log|("u" - 3)/("u" + 3)| + "C"_2`

= `1/12 log|(x^2 - 3)/(x^2 + 3)| + "C"_2`.

Thus I = I₁ + I₂

= `1/4 log|x^4 - 9| + 1/12 log|(x^2 - 3)/(x^2 + 3)| + "C"`

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Definite Integrals

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Chapter 7: Integrals - Solved Examples [Page 154]

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NCERT Exemplar Mathematics [English] Class 12

Chapter 7 Integrals
Solved Examples | Q 16 | Page 154

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