Question

please provide the solution for this question asap​

Answer 1

Answer--->

( 1 / 9 ) tan⁻¹ ( x³ - 2 ) / 3 + C

Step-by-step explanation:

To find---> ∫ x² dx / ( x⁶ - 4 x³ + 13 )

Solution---> Let,

I = ∫ x² dx / ( x⁶ - 4 x³ + 13 )

= ∫ x² dx / { ( x³ )² - 4 ( x³ ) + 13 }

Let , x³ = t

Differentiating with respect to x , we get,

=> 3 x² dx = dt

=> x² dx = dt / 3

Now,

I = ∫ dt / 3 ( t² - 4 t + 13 )

= ∫ dt / 3 { ( t² - 4t + 4 ) + 9 }

= ∫ dt / 3 { ( t - 2 )² + ( 3 )² }

We know that,

∫ dy /( y² + a² ) = 1 / a tan⁻¹ ( y / a ) + C , using it here , we get

= ( 1 / 3 ) ( 1 / 3 ) tan⁻¹ ( t - 2 ) / 3 + C

= ( 1 / 9 ) tan⁻¹ ( x³ - 2 ) / 3 + C