Здравствуйте, Aspir!
1. =1/7^0.5arctg(x/7^0/5)+c
2.=1/2*((2x+3)^15)/15+c=1/30*(2x+3)^15+c

Здравствуйте, Aspir.

1. ∫dx/(7 + x²) = ∫dx/((√7)² + x²) = 1/√7 ∙ arctg x/√7 + C.

2. ∫(2x + 3)¹⁴dx = ∫(2x + 3)¹⁴ ∙ 1/2 ∙ d(2x + 3) = 1/2 ∙ ∫(2x + 3)¹⁴ ∙ d(2x + 3) =
= 1/2 ∙ 1/15 ∙ (2x + 3)¹⁵ + C = 1/30 ∙ (2x + 3)¹⁵ + C.

3. ∫(3x + 2)dx/(x² – 2x + 3) = ∫(3/2 ∙ (2x – 2) + 5)dx/(x² – 2x + 3) =
= 3/2 ∙ ∫(2x – 2)dx/(x² – 2x + 3) + 5∫dx/((x – 1)² + (√2)²) =
= 3/2 ∙ ∫d(x² – 2x + 3)/(x² – 2x + 3) + 5∫d(x – 1)/((x – 1)² + (√2)²) =
3/2 ∙ ln (x² – 2x + 3) + 5/√2 ∙ arctg (x – 1)/√2 + C.

4. ∫(2x – 5)dx/√(x² + 25) = ∫2xdx/√(x² + 25) + ∫5dx/√(x² + 25) =
= ∫d(x² + 25)/√(x² + 25) + 5∫dx/√(x² + 5²) = √(x² + 25)/(1/2) + 5 ∙ ln |x + √(x² + 25)| + C =
= 2√(x² + 25) + 5 ∙ ln |x + √(x² + 25)| + C.

5. ∫arcsin⁵ 2x ∙ dx/√(1 – 4x²) = ∫arcsin⁵ 2x ∙ 1/2 ∙ d(2x)/√(1 – (2x)²) = 1/2 ∙ ∫arcsin⁵ 2x ∙ d(arcsin 2x) =
= 1/6 ∙ arcsin⁶ 2x + C.

6. ∫e^3xdx/(2e^3x – 5) = ∫1/6 ∙ d(2e^3x – 5)/(2e^3x – 5) = 1/6 ∙ ln |2e^3x – 5| + C.

7. ∫x³dx/√(x⁴ – 9) = ∫1/4 ∙ 4x³dx/√(x⁴ – 9) = 1/4 ∙ ∫d(x4 – 9)/√(x⁴ – 9) = 1/4 ∙ √(x⁴ – 9)/(1/2) + C₁ =
= 1/2 ∙ √(x⁴ – 9) + C₁,
∫xdx/√(x⁴ – 9) = ∫1/2 ∙ d(x²)/√((x²)² – 9) = 1/2 ∙ ∫d(x²)/√((x²)² – 9) = 1/2 ∙ ln |x² + √(x⁴ – 9)| + C₂,

∫(3x³ – x)dx/√(x⁴ – 9) = 3∫x³dx/√(x⁴ – 9) - ∫xdx/√(x⁴ – 9) = 3/2 ∙ √(x⁴ – 9) – 1/2 ∙ ln |x² + √(x⁴ – 9)| + C.

8. ∫cos 4x ∙ dx = ∫cos 4x ∙ 1/4 ∙ d(4x) = 1/4 ∙ ∫cos 4x ∙ d(4x) = -1/4 ∙ sin 4x + C₁,
∫x ∙ cos 4x ∙ dx = (u = x, dv = cos 4x ∙ dx, du = dx, v = -1/4 ∙ sin 4x) = -1/4 ∙ x ∙ sin 4x + 1/4 ∙ ∫sin 4x ∙ dx =
= -1/4 ∙ x ∙ sin 4x + 1/4 ∙ 1/4 ∙ cos 4x + C₂ = -1/4 ∙ x ∙ sin 4x + 1/16 ∙ cos 4x + C₂,

∫(2x – 1)cos 4x ∙ dx = ∫2x ∙ cos 4x ∙ dx - ∫cos 4x ∙ dx = 2∫x ∙ cos 4x ∙ dx - ∫cos 4x ∙ dx =
= -1/2 ∙ x ∙ sin 4x + 1/8 ∙ cos 4x + 1/4 ∙ sin 4x + C.

9. ∫ln 3x ∙ dx = (u = ln 3x, dv = dx, du = 3dx/(3x) = dx/x, v = x) = x ∙ ln 3x - ∫dx = x ∙ ln 3x – x + C₁,
∫x ∙ ln 3x ∙ dx = (u = ln 3x, dv = xdx, du = dx/x, v = x²/2) = 1/2 ∙ x² ∙ ln 3x – 1/2 ∙ ∫xdx =
= 1/2 ∙ x² ∙ ln 3x – 1/4 ∙ x² + C₂,

∫(x + 1)ln 3x ∙ dx = ∫x ∙ ln 3x ∙ dx + ∫ln 3x ∙ dx = 1/2 ∙ x² ∙ ln 3x – 1/4 ∙ x² + x ∙ ln 3x – x + C.

10. ∫x ∙ arcsin x ∙ dx/√(1 – x²) = (x = sin t, arcsin x = t, dx/√(1 – x²) = dt) = ∫t ∙ sin t ∙ dt =
= (u = t, dv = sin t ∙ dt, du = dt, v = -cos t) = -t ∙ cos t + ∫cos t ∙ dt = -t ∙ cos t – sin t + C =
= -arcsin x ∙ cos arcsin x – x + C = -√(1 – x²) ∙ arcsin x – x + C.

11. ∫(x⁴ – 5x³ + 7x² – 4x + 2)dx/((x(x – 1)(x – 2)) = ∫(x⁴ – 5x³ + 7x² – 4x + 2)dx/(x³ – 3x² + 2x) =
= ∫(x – 2 – (x² – 2)/(x³ – 3x² + 2x))dx = ∫xdx - 2∫dx - ∫(x² – 2)dx/(x³ – 3x² + 2x) =
= ∫xdx - 2∫dx - ∫(3x² – 6x + 2 – 2x² + 6x – 4)dx/(x³ – 3x² + 2x) =
= ∫xdx - 2∫dx - ∫(3x² – 6x + 2)dx/(x³ – 3x² + 2x) – ∫(2x² – 6x + 4)dx/(x³ – 3x² + 2x) =
= ∫xdx - 2∫dx - ∫(3x² – 6x + 2)dx/(x³ – 3x² + 2x) – 2∫(x² – 3x + 2)dx/(x(x² – 3x + 2)) =
= ∫xdx - 2∫dx - ∫d(x³ – 3x² + 2x)/(x³ – 3x² + 2x) – 2∫dx/x =
= 1/2 ∙ x² – ln |x³ – 3x² + 2x| - 2ln |x| + C.

12. 9/(x²(x² + 9)) = (x² + 9 – x²)/(x²(x² + 9)) = 1/x² – 1/(x² + 9),

∫(-2x³ + 2x² – 18x + 9)dx/(x²(x² + 9)) = ∫(-2x(x² + 9)dx/(x²(x² + 9)) + ∫(2x² + 9)dx/(x²(x² + 9)) =
= -2∫dx/x + ∫2dx/(x² + 9) + 9∫dx/(x²(x² + 9)) = -2∫dx/x + ∫2dx/(x² + 9) + 9∫dx/x² - 9∫dx/(x² + 9) =
= -2∫dx/x + ∫d(x² + 9)/(x² + 9) + 9∫dx/x² - 9∫dx/(x² + 9) =
= -x² + ln (x² + 9) – 9/x – 3 ∙ arctg x/3 + C.

13. ∫dx/(1 + sin x) = (t = tg x/2, sin x = 2t/(1 + t²), dx = 2dt/(1 + t²)) = ∫2dt/(1 + t²)) ∙ 1/(1 + 2t/(1 + t²)) =
= 2∫dt/(1 + t²) ∙ (1 + t²)/(1 + t)² = 2∫dt/(1 + t)² = 2∫d(1 + t)/(1 + t)² = -2/(1 + t) + C.

14. t²/((1 – t)(1 + t²)) = A/(1 – t) + (Bt + C)/(1 + t²) = (A(1 + t²) + (Bt + C)(1 – t))/((1 – t)(1 + t²)),
A(1 + t²) + (Bt + C)(1 – t) = t²,
(A – B)t² + (B – C)t + A + C = t²,
A – B = 1,
B – C = 0, A + C = 0,
A – B = 1, A + B = 0,
A = 1/2, B = -1/2, C = -1/2,
t²/((1 – t)(1 + t²)) = 1/(2(1 – t)) - (t + 1)/(2(1 + t²)),

∫tg² x ∙ dx/(1 – tg x) = (x = arctg t, dx = dt/(1 + t²)) = ∫t² ∙ dt/(1 + t²) ∙ 1/(1 – t) = ∫t²dt/((1 – t)(1 + t²)) =
= 1/2 ∙ ∫dt/(1 – t) – 1/2 ∙ ∫tdt/(1 + t²) – 1/2 ∙ ∫dt/(1 + t²) =
= -1/2 ∙ ∫d(1 – t)/(1 – t) – 1/4 ∙ ∫d(1 + t²)/(1 + t²) – 1/2 ∙ ∫dt/(1 + t²) =
= -1/2 ∙ ln |1 – t| - 1/4 ∙ ln (1 + t²) - 1/2 ∙ arctg t + C₁ =
= -1/2 ∙ ln |1 – tg x| - 1/4 ∙ ln (1 + tg² x) - x/2 + C₁,

∫(2tg² x – tg x + 1)dx/(1 – tg x) = 2∫tg² x ∙ dx/(1 – tg x) + ∫(1 – tg x)dx/(1 – tg x) =
= 2∫tg² x ∙ dx/(1 – tg x) + ∫dx = -ln |1 – tg x| - 1/2 ∙ ln (1 + tg² x) – x + x + C =
= -ln |1 – tg x| - 1/2 ∙ ln (1 + tg² x) + C.

Что касается 15-го интеграла, то Вам необходимо уточнить структуру подынтегрального выражения. И не следует в одном вопросе помещать столь много заданий.

С уважением.