Đáp án:

$a)\ \ e^x\cdot \tan\dfrac{x}{2} + C$

$b)\ \ \dfrac{1}{54}(x+1)^2(4x + 31) + \dfrac13(x+1)^2(x+4)\ln^2(x+1) - \dfrac19(x+1)^2(2x+11)\ln(x+1) + C$

Giải thích các bước giải:

\(\begin{array}{l}
a)\quad I = \displaystyle\int e^x\dfrac{1+\sin x}{1+\cos x}dx\\
\Leftrightarrow I = \displaystyle\int e^x \dfrac{1 + 2\sin\dfrac{x}{2}\cdot\cos\dfrac{x}{2}}{1 + \left(2\cos^2\dfrac{x}{2} - 1\right)}dx\\
\Leftrightarrow I = \displaystyle\int e^x\left(\dfrac{2\sin\dfrac{x}{2}\cdot \cos\dfrac{x}{2}}{2\cos^2\dfrac{x}{2}} + \dfrac{1}{2\cos^2\dfrac{x}{2}}\right)dx\\
\Leftrightarrow I = \displaystyle\int\left(e^x\cdot \tan\dfrac{x}{2} + e^x\cdot \dfrac{1}{2\cos^2\dfrac{x}{2}}\right)dx\\
\Leftrightarrow I = \displaystyle\int \left[ (e^x)'\cdot \tan\dfrac{x}{2} + e^x \cdot \left(\tan\dfrac{x}{2}\right)'\right]dx\\
\Leftrightarrow I = \displaystyle\int \left(e^x\cdot \tan\dfrac{x}{2}\right)'dx\\
\Leftrightarrow I = e^x\cdot \tan\dfrac{x}{2} + C
\end{array}\)

$b)\quad I = \displaystyle\int (x^2 + 4x + 3)\ln^2(x+1)dx$

$\Leftrightarrow I = \displaystyle\int (x+3)(x+1)\ln^2(x+1)dx$

Đặt $\begin{cases}u = x+3\\dv = (x+1)\ln^2(x+1)dx\end{cases}\Rightarrow \begin{cases}du = dx\\v = \displaystyle\int (x+1)\ln^2(x+1)dx\end{cases}$

Xét $I' = \displaystyle\int (x+1)\ln^2(x+1)dx$

$\Leftrightarrow I' =  \dfrac12(x+1)^2\ln^2(x+1) - \displaystyle\int \dfrac{(x+1)^2}{2}d[\ln^2(x+1)]$

$\Leftrightarrow I' = \dfrac12(x+1)^2\ln^2(x+1) - \displaystyle\int (x+1)\ln(x+1)dx$

$\Leftrightarrow I' = \dfrac12(x+1)^2\ln^2(x+1) - \left[\dfrac12(x+1)^2\ln(x+1) - \displaystyle\int \dfrac{(x+1)^2}{2}d[\ln(x+1)]\right]$

$\Leftrightarrow I' = \dfrac12(x+1)^2\ln^2(x+1) - \dfrac12(x+1)^2\ln(x+1) + \dfrac14(x+1)^2 + C_0$

Do đó, chọn

$v = \dfrac12(x+1)^2\ln^2(x+1) - \dfrac12(x+1)^2\ln(x+1) + \dfrac14(x+1)^2$

Khi đó:

$\quad I = \dfrac12(x+1)^2(x+3)\ln^2(x+1) - \dfrac12(x+1)^2(x+3)\ln(x+1) + \dfrac14(x+1)^2(x+3) - \displaystyle\int\left[\dfrac12(x+1)^2\ln^2(x+1) - \dfrac12(x+1)^2\ln(x+1) + \dfrac14(x+1)^2\right]dx$

Hoàn toàn tương tự như trên, ta được:

$\displaystyle\int \dfrac12(x+1)^2\ln^2(x+1)dx = \dfrac{1}{27}(x+1)^3 + \dfrac16(x+1)^3\ln^2(x+1) - \dfrac19(x+1)^3\ln(x+1) + C_1$

$\displaystyle\int \dfrac12(x+1)^2\ln(x+1)dx = - \dfrac{1}{18}(x+1)^3 + \dfrac16(x+1)^3\ln(x+1)+C_2$

$\displaystyle\int \dfrac14(x+1)^2dx = \dfrac{1}{12}(x+1)^3 + C_3$

Thay vào $I$ và rút gọn, ta được:

$I = \dfrac{1}{54}(x+1)^2(4x + 31) + \dfrac13(x+1)^2(x+4)\ln^2(x+1) - \dfrac19(x+1)^2(2x+11)\ln(x+1) + C$

Đáp án:

Giải thích các bước giải:

Câu a) thì cậu ấy tính OK rồi, tớ mạn phép có ý kiến

về câu b) chút . Nhận xét:

$ I = \int\limits{(x^{2} + 4x + 3)ln^{2}(x + 1)} \, dx $

$ = \int\limits{(x + 1)^{2}ln^{2}(x + 1)} \, dx +  2\int\limits{(x + 1)ln^{2}(x + 1)} \, dx$
Đặt $: t = ln(x + 1) => x + 1 = e^{t}; dx = (x + 1)dt $

$ J =  \int\limits{(x + 1)^{2}ln^{2}(x + 1)} \, dx $

$  = \int\limits{t^{2}e^{3t}} \, dt = \dfrac{1}{3}\int\limits{t^{2}} \,d(e^{3t})$

$ = \dfrac{1}{3}t^{2}e^{3t} - \dfrac{2}{9}\int\limits{t} \, d(e^{3t})$

$ = \dfrac{1}{3}t^{2}e^{3t} - \dfrac{2}{9}te^{3t} + \dfrac{2}{9}\int\limits{e^{3t}} \, dt$
$ = \dfrac{1}{3}t^{2}e^{3t} - \dfrac{2}{9}te^{3t} + \dfrac{2}{27}e^{3t} + C_{1}$

$ K =  \int\limits{(x + 1)ln^{2}(x + 1)} \, dx $

$  = \int\limits{t^{2}e^{2t}} \, dt = \dfrac{1}{2}\int\limits{t^{2}} \,d(e^{2t})$

$ = \dfrac{1}{2}t^{2}e^{2t} - \dfrac{1}{2}\int\limits{t} \, d(e^{2t})$

$ = \dfrac{1}{3}t^{2}e^{3t} - \dfrac{1}{2}te^{2t} + \dfrac{1}{2}\int\limits{e^{2t}} \, dt$

$ = \dfrac{1}{2}t^{2}e^{2t} - \dfrac{1}{2}te^{2t} + \dfrac{1}{4}e^{2t} + C_{2}$

$ I = J + 2K = \dfrac{1}{54}e^{2t}[18(e^{t} + 3)t^{2} - 6(2e^{t} + 9)t + (4e^{t} + 27)] + C$

$ = \dfrac{1}{54}(x + 1)^{2}[18(x + 4)ln^{2}(x + 1) - 6(2x + 11)ln(x + 1) + 4x + 31] + C$