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Đặt \(\int f\left(x\right)dx=F\left(x\right)\Rightarrow\int\limits^{17}_1f\left(x\right)dx=F\left(17\right)-F\left(1\right)\)
Từ giả thiết:
\(2x.f\left(x^2+1\right)+\dfrac{f\left(\sqrt{x}\right)}{2\sqrt{x}}=2lnx\)
Lấy nguyên hàm 2 vế:
\(F\left(x^2+1\right)+F\left(\sqrt{x}\right)=2xlnx-2x+C\)
Thay \(x=4\):
\(F\left(17\right)+F\left(2\right)=16ln2-8+C\) (1)
Thay \(x=1\):
\(F\left(2\right)+F\left(1\right)=-2+C\) (2)
Trừ vế cho vế (1) cho (2):
\(F\left(17\right)-F\left(1\right)=16ln2-6\)
Vậy \(\int\limits^{17}_1f\left(x\right)dx=16ln2-6\)
G
ồm 
Câu 17:
\(F(x)=\int \sqrt{\ln^2x+1}\frac{\ln x}{x}dx=\int \sqrt{\ln ^2x+1}\ln xd(\ln x)\)
\(\Leftrightarrow F(x)=\frac{1}{2}\int \sqrt{\ln ^2x+1}d(\ln ^2x)\)
Đặt \(\sqrt{\ln^2 x+1}=t\) \(\Rightarrow \ln ^2x=t^2-1\)
\(\Rightarrow F(x)=\frac{1}{2}\int td(t^2-1)=\int t^2dt=\frac{t^3}{3}+c=\frac{\sqrt{(\ln^2x+1)^3}}{3}+c\)
Vì \(F(1)=\frac{1}{3}\Leftrightarrow \frac{1}{3}+c=\frac{1}{3}\Rightarrow c=0\)
\(\Rightarrow F^2(e)=\left(\frac{\sqrt{\ln ^2e+1)^3}}{3}\right)^2=\frac{8}{9}\)
Câu 11)
Đặt \(\sqrt{3x+1}=t\Rightarrow x=\frac{t^2-1}{3}\)
\(\Rightarrow I=\int ^{5}_{1}\frac{dx}{x\sqrt{3x+1}}==\int ^{5}_{1}\frac{d\left ( \frac{t^2-1}{3} \right )}{\frac{t(t^2-1)}{3}}=\int ^{4}_{2}\frac{2tdt}{t(t^2-1)}=\int ^{4}_{2}\frac{2dt}{(t-1)(t+1)}\)
\(=\int ^{4}_{2}\left ( \frac{dt}{t-1}-\frac{dt}{t+1} \right )=\left.\begin{matrix} 4\\ 2\end{matrix}\right|(\ln|t-1|-\ln|t+1|)=2\ln 3-\ln 5\)
\(\Rightarrow a=2,b=-1\Rightarrow a^2+ab+3b^2=5\)
Đáp án C
Câu 20)
Ta có:
\(I=\int ^{x}_{\frac{1}{e}}\frac{\ln t+1}{t}dt=\int ^{x}_{\frac{1}{e}}(\ln t+1)d(\ln t)=\int ^{x}_{\frac{1}{e}}\ln td(\ln t)+\int ^{x}_{\frac{1}{e}}d(\ln t)\)
\(=\left.\begin{matrix} x\\ \frac{1}{e}\end{matrix}\right|\left ( \ln t+\frac{\ln^2t}{2}+c \right )=\left ( \ln x+\frac{\ln^2x}{2} \right )+\frac{1}{2}=18\leftrightarrow \ln x+\frac{\ln ^2x}{2}=\frac{35}{2}\)
\(\Rightarrow\left[\begin{matrix}x=e^{-7}\\x=e^5\end{matrix}\right.\)
Đáp án A.
. Em cảm ơn
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