For I_n = ∫_0^1 (x^n)/(1 + x) dx, the sum I_n + I_n+1 equals:

I_n + I_n+1 = ∫_0^1 dfracx^n + x^n+11 + x dx = ∫_0^1 x^n dx = (1)/(n + 1).

Daily Math Problems — June 11, 2026: Logarithms, Powers, Radicals, Logarithms, Calculus

Problems & worked solutions

RO M1

$\frac{\log_{3} 81}{\log_{3} 9}$ equals:

Show answer & worked solution

$\log_{3} 81 = 4$ , $\log_{3} 9 = 2$ , so the ratio is $\frac{4}{2} = 2$ .

RO M1

If $x = \log_{2} 5$ , then $4^{x}$ equals:

Show answer & worked solution

$4^{x} = 2^{2 x} = (2^{x})^{2} = 5^{2} = 25$ .

RO M1

For $I_{n} = \int_{0}^{1} \frac{x^{n}}{1 + x} d x$ , the sum $I_{n} + I_{n + 1}$ equals:

Show answer & worked solution

$I_{n} + I_{n + 1} = \int_{0}^{1} \frac{x^{n} + x^{n + 1}}{1 + x} d x = \int_{0}^{1} x^{n} d x = \frac{1}{n + 1}$ .

Problems & worked solutions

RO M1

$\frac{\log_{3} 81}{\log_{3} 9}$ equals:

Show answer & worked solution

$\log_{3} 81 = 4$ , $\log_{3} 9 = 2$ , so the ratio is $\frac{4}{2} = 2$ .

RO M1

If $x = \log_{2} 5$ , then $4^{x}$ equals:

Show answer & worked solution

$4^{x} = 2^{2 x} = (2^{x})^{2} = 5^{2} = 25$ .

RO M1

For $I_{n} = \int_{0}^{1} \frac{x^{n}}{1 + x} d x$ , the sum $I_{n} + I_{n + 1}$ equals:

Show answer & worked solution

$I_{n} + I_{n + 1} = \int_{0}^{1} \frac{x^{n} + x^{n + 1}}{1 + x} d x = \int_{0}^{1} x^{n} d x = \frac{1}{n + 1}$ .