Tính

Để tính nội hàm \(\int_0^{\frac{\pi}{4}} (2 \tan^2 x + 5) \, dx\), ta tách nó thành hai phần:

\[
\int_0^{\frac{\pi}{4}} (2 \tan^2 x + 5) \, dx = \int_0^{\frac{\pi}{4}} 2 \tan^2 x \, dx + \int_0^{\frac{\pi}{4}} 5 \, dx
\]

**Bước 1: Tính \(\int_0^{\frac{\pi}{4}} 5 \, dx\)**

\[
\int_0^{\frac{\pi}{4}} 5 \, dx = 5 \left[ x \right]_0^{\frac{\pi}{4}} = 5 \left( \frac{\pi}{4} - 0 \right) = \frac{5\pi}{4}
\]

**Bước 2: Tính \(\int_0^{\frac{\pi}{4}} 2 \tan^2 x \, dx\)**

Ta biết rằng:

\[
\tan^2 x = \sec^2 x - 1
\]

Vậy:

\[
\int_0^{\frac{\pi}{4}} 2 \tan^2 x \, dx = 2 \int_0^{\frac{\pi}{4}} (\sec^2 x - 1) \, dx = 2 \left( \int_0^{\frac{\pi}{4}} \sec^2 x \, dx - \int_0^{\frac{\pi}{4}} 1 \, dx \right)
\]

- Tính \(\int_0^{\frac{\pi}{4}} \sec^2 x \, dx\):

\[
\int \sec^2 x \, dx = \tan x \Rightarrow \left[ \tan x \right]_0^{\frac{\pi}{4}} = \tan\left(\frac{\pi}{4}\right) - \tan(0) = 1 - 0 = 1
\]

- Tính \(\int_0^{\frac{\pi}{4}} 1 \, dx\):

\[
\int_0^{\frac{\pi}{4}} 1 \, dx = \frac{\pi}{4}
\]

Vậy:

\[
\int_0^{\frac{\pi}{4}} 2 \tan^2 x \, dx = 2 \left( 1 - \frac{\pi}{4} \right) = 2 - \frac{\pi}{2}
\]

**Bước 3: Cộng các kết quả lại**

\[
\int_0^{\frac{\pi}{4}} (2 \tan^2 x + 5) \, dx = \left( 2 - \frac{\pi}{2} \right) + \frac{5\pi}{4}
\]

\[
= 2 - \frac{\pi}{2} + \frac{5\pi}{4} = 2 + \left(-\frac{2\pi}{4} + \frac{5\pi}{4}\right) = 2 + \frac{3\pi}{4}
\]

Cuối cùng, kết quả là:

\[
\int_0^{\frac{\pi}{4}} (2 \tan^2 x + 5) \, dx = 2 + \frac{3\pi}{4}
\]