1/(x+10)(x+11)+1/(x+11)(x+12)+1/(x+12)(x+13)+1/x 13=1/14

1/(x+10)(x+11)+1/(x+11)(x+12)+1/(x+12)(x+13)+1/x 13=1/14

1/(x+10)(x+11)+1/(x+11)(x+12)+1/(x+12)(x+13)+1/(x+13)=1/14
∴1/(x+10)-1/(x+11)+1/(x+11)-1/(x+12)+1/(x+12)-1/(x+13)+1/(x+13)=1/14
∴1/(x+10)=1/14
∴x=4

Find the intersection of the functions y = log10 (X &; 2) and y = 1 &; log10 (x + 1). Why are X-2 and X + 1 positive numbers?

A:
Find the intersection of the functions y = log10 (X &; 2) and y = 1 &; log10 (x + 1). Why are X-2 and X + 1 positive numbers
Because the true number of logarithmic function must be positive
So:
x-2>0
x+1>0
Otherwise, the logarithmic function is meaningless
y = log10(x − 2) = 1 − log10(x + 1)=log10[10/(x+1)]
x-2=10/(x+1)
x^2-x-2=10
x^2-x-12=0
(x-4)(x+3)=0
X = 4 or x = - 3
Because: X-2 > 0, x + 1 > 0
So: x > 2
So: x = - 3 does not match
To sum up, x = 4, y = log10 (2)
So: the intersection point is (4, LG2)

Can the great God tell me how to find ∫ (1 + x ^ 4) / (1 + x ^ 6) DX

x⁶ + 1 = x²[(x⁴ + 1) - 1] + 1
= x²(x⁴ + 1) - x² + 1
∫ (x⁶ + 1)/(x⁴ + 1) dx
= ∫ [x²(x⁴ + 1) + 1 - x²]/(x⁴ + 1) dx
= ∫ x² dx - ∫ (x² - 1)/(x⁴ + 1) dx
= x³/3 - ∫ (1 - 1/x²)/(x² + 1/x²) dx
= x³/3 - ∫ d(x + 1/x)/[(x + 1/x)² - 2]
= x³/3 - 1/(2√2) * ln| [(x + 1/x) - √2]/[(x + 1/x) + √2] | + C
= x³/3 - (√2/4)ln| (x² - √2x + 1)/(x² + √2x + 1) | + C