第12类RD Sharma解–第19章不定积分–练习19.15

问题1.评估∫1 /(4x ² + 12x + 5)dx

解决方案：

Let I = ∫ 1/(4x² + 12x + 5) dx

by taking 1/4 common from the above eq

= 1/4 ∫ 1/ x² + 3x + 5/4 dx

= 1/4 ∫ 1/ x² + 2x × (3/2)x + (3/2)² – (3/2)² + 5/4 dx

= 1/4 ∫ 1/ (x + 3/2)² – 1 dx (i)

put (x+ 3/2) = t

dx = dt

put the above value in eq. (i)

= 1/4 ∫ 1/ t² – (1)² dt

Integrate the above eq. then, we get

= 1/4 × 1/2×(1) log |t-1/t+1| +c [since, ∫1/x² – a² dx = 1/2a log|x-a/x+a| +c]

put the value of t in the above eq.

= 1/8 log|x+ 3/2 – 1/x+ 3/2 + 1| + c

Hence, I = 1/8 log|2x+1/ 2x+5| + c

问题2.评估∫1/ x ² – 10x + 34 dx

解决方案：

Let I = ∫1/x² – 10x + 34 dx

=∫1/x² – 2x × 5 + (5)² – (5)² + 34 dx

=∫1/ (x – 5)² + 9 dx (i)

substituting (x-1) = t

dx = dt

put the above value in eq. (i)

= ∫ 1/ t² + (3)² dt

Integrate the above eq. then, we get

= 1/3 tan^-1 (t/3) + c [Since, ∫ 1/x² + a² dx = 1/a tan^-1 (x/2) + c]

Put the value of t in the above eq.

Hence, I = 1/3 tan^-1 (x-5/ 3) + c

问题3.评估∫1 / 1-xx ² dx

解决方案：

Let I = ∫ 1/ 1-x-x² dx

= ∫ 1/ -(x² – x – 1) dx

adding and subtracting 1/4 in the denominator to make it a perfect square

= ∫ 1/ -(x² – x + 1/4 – 1 – 1/4) dx

= ∫ 1/ -([x² – x + 1/4] – 1 – 1/4) dx

= ∫ 1/ -([x – 1/2]² – 5/4) dx

= ∫ 1/ (5/4 – [x – 1/2]²) dx

= ∫ 1/ ([√5/2]² – [x – 1/2]²) dx

Integrate the above eq. then, we get

= 1/2(√5/2) log|√5/2 + (x-1/2)/ √5/2 – (x-1/2)| + c [since ∫ 1/a² + x² dx = 1/2a log|x+a/x-a| +c]

Hence, I = 1/√5 log|√5/2 + (x-1/2) /√5/2 – (x-1/2)| + c

问题4.评估∫1 / 2x ² – x – 1 dx

解决方案：

Let I = ∫ 1/2x² – x – 1 dx

taking 1/2 common from the above eq.

=1/2 ∫ 1/ x² – x/2 – 1/2 dx

=1/2 ∫ 1/ x² – 2x × 1/4 + (1/4)² – (1/4)² – 1/2 dx

= 1/2 ∫ 1/ (x – 1/4)² – 9/16 dx

put, x- 1/4 = t

dx = dt

= 1/2 ∫ 1/ t² – (3/4)² dt

Integrate the above eq. then, we get

= (1/2) 1/[2×(3/4)] log|t-(3/4) / t+(3/4)| + c [Since, ∫ 1/x² -a² dx = 1/2a log|x – a/ x+a| + c]

Put the value of t in above eq.

= 1/3 log|(x-1/4-3/4)/x-1/4+3/4| + c

Hence, I = 1/3log|x – 1/2x+1| + c

问题5.评估∫dx / x ² + 6x +13

解决方案：

Let I =∫ dx/x² + 6x +13 (i)

We have x² + 6x +13 = x² + 6x + 3² – 3² +13 =(x + 3)² + 4

Put the above value in eq. (i)

∫ 1/x² + 6x +13 dx = ∫ 1/(x+3)² + 2² dx

put x+3 = t and

dx = dt

= ∫ dt/ t² + 2²

Integrate the above eq. then, we get

= 1/2 tan^-1 t/2 + c

Put the value of t in above eq.

Hence, I = 1/2 tan^-1 x+3/2 + c