\(\int\left(3x^2-2x-4\right)dx=x^3-x^2-4x+C\)

\(\int\left(sin3x-cos4x\right)dx=-\dfrac{1}{3}cos3x-\dfrac{1}{4}sin4x+C\)

\(\int\left(e^{-3x}-4^x\right)dx=-\dfrac{1}{3}e^{-3x}-\dfrac{4^x}{ln4}+C\)

d. \(I=\int lnxdx\)

Đặt \(\left\{{}\begin{matrix}u=lnx\\dv=dx\end{matrix}\right.\) \(\Rightarrow\left\{{}\begin{matrix}du=\dfrac{dx}{x}\\v=x\end{matrix}\right.\)

\(\Rightarrow u=x.lnx-\int dx=x.lnx-x+C\)

e. Đặt \(\left\{{}\begin{matrix}u=x\\dv=e^xdx\end{matrix}\right.\) \(\Rightarrow\left\{{}\begin{matrix}du=dx\\v=e^x\end{matrix}\right.\)

\(\Rightarrow I=x.e^x-\int e^xdx=x.e^x-e^x+C\)

f.

Đặt \(\left\{{}\begin{matrix}u=x+1\\dv=sinxdx\end{matrix}\right.\) \(\Rightarrow\left\{{}\begin{matrix}du=dx\\v=-cosx\end{matrix}\right.\)

\(\Rightarrow I=-\left(x+1\right)cosx+\int cosxdx=-\left(x+1\right)cosx+sinx+C\)

g.

Đặt \(\left\{{}\begin{matrix}u=lnx\\dv=xdx\end{matrix}\right.\) \(\Rightarrow\left\{{}\begin{matrix}du=\dfrac{dx}{x}\\v=\dfrac{1}{2}x^2\end{matrix}\right.\)

\(\Rightarrow I=\dfrac{1}{2}x^2.lnx-\dfrac{1}{2}\int xdx=\dfrac{1}{2}x^2.lnx-\dfrac{1}{4}x^2+C\)

a) \(\int\left(x+\ln x\right)x^2\text{d}x=\int x^3\text{d}x+\int x^2\ln x\text{dx}\)

\(=\dfrac{x^4}{4}+\int x^2\ln x\text{dx}+C\) (*)

Để tính: \(\int x^2\ln x\text{dx}\) ta sử dụng công thức tính tích phân từng phần như sau:

Đặt \(\left\{{}\begin{matrix}u=\ln x\\v'=x^2\end{matrix}\right.\) \(\Rightarrow\left\{{}\begin{matrix}u'=\dfrac{1}{x}\\v=\dfrac{1}{3}x^3\end{matrix}\right.\)

Suy ra:

\(\int x^2\ln x\text{dx}=\dfrac{1}{3}x^3\ln x-\dfrac{1}{3}\int x^2\text{dx}\)

\(=\dfrac{1}{3}x^3\ln x-\dfrac{1}{3}.\dfrac{1}{3}x^3\)

Thay vào (*) ta tính được nguyên hàm của hàm số đã cho bằng:

(*) \(=\dfrac{1}{3}x^3-\dfrac{1}{3}x^3\ln x+\dfrac{1}{9}x^3+C\)

\(=\dfrac{4}{9}x^3-\dfrac{1}{3}x^3\ln x+C\)

b) Đặt \(\left\{{}\begin{matrix}u=x+\sin^2x\\v'=\sin x\end{matrix}\right.\)

\(\Rightarrow\left\{{}\begin{matrix}u'=1+2\sin x.\cos x\\v=-\cos x\end{matrix}\right.\)

Ta có:

\(\int\left(x+\sin^2x\right)\sin x\text{dx}=-\left(x+\sin^2x\right)\cos x+\int\left(1+2\sin x\cos^2x\right)\text{dx}\)

\(=-\left(x+\sin^2x\right)\cos x+\int\cos x\text{dx}+2\int\sin x.\cos^2x\text{dx}\)

\(=-\left(x+\sin^2x\right)\cos x+\sin x-2\int\cos^2x.d\left(\cos x\right)\)

\(=-\left(x+\sin^2x\right)\cos x+\sin x-2\dfrac{\cos^3x}{3}+C\)

c) Đặt \(\left\{{}\begin{matrix}u=x+e^x\\v'=e^{2x}\end{matrix}\right.\) \(\Rightarrow\left\{{}\begin{matrix}u'=1+e^x\\v=\dfrac{1}{2}e^{2x}\end{matrix}\right.\)

Ta có:

\(\int\left(x+e^x\right)e^{2x}\text{dx}=\dfrac{1}{2}\left(x+e^x\right)e^{2x}-\dfrac{1}{2}\int\left(1+e^x\right)e^{2x}\text{dx}\)

\(=\dfrac{1}{2}\left(x+e^x\right)e^{2x}-\dfrac{1}{2}\int e^{2x}\text{dx}-\dfrac{1}{2}\int e^{3x}\text{dx}\)

\(=\dfrac{1}{2}\left(x+e^x\right)e^{2x}-\dfrac{1}{2}.\dfrac{1}{2}e^{2x}-\dfrac{1}{2}.\dfrac{1}{3}e^{3x}\)

\(=\dfrac{1}{2}xe^{2x}-\dfrac{1}{4}e^{2x}+\dfrac{1}{3}e^{3x}\)

a)

Ta có \(A=\int ^{\frac{\pi}{4}}_{0}\cos 2x\cos^2xdx=\frac{1}{4}\int ^{\frac{\pi}{4}}_{0}\cos 2x(\cos 2x+1)d(2x)\)

\(\Leftrightarrow A=\frac{1}{4}\int ^{\frac{\pi}{2}}_{0}\cos x(\cos x+1)dx=\frac{1}{4}\int ^{\frac{\pi}{2}}_{0}\cos xdx+\frac{1}{8}\int ^{\frac{\pi}{2}}_{0}(\cos 2x+1)dx\)

\(\Leftrightarrow A=\frac{1}{4}\left.\begin{matrix} \frac{\pi}{2}\\ 0\end{matrix}\right|\sin x+\frac{1}{16}\left.\begin{matrix} \frac{\pi}{2}\\ 0\end{matrix}\right|\sin 2x+\frac{1}{8}\left.\begin{matrix} \frac{\pi}{2}\\ 0\end{matrix}\right|x=\frac{1}{4}+\frac{\pi}{16}\)

b)

\(B=\int ^{1}_{\frac{1}{2}}\frac{e^x}{e^{2x}-1}dx=\frac{1}{2}\int ^{1}_{\frac{1}{2}}\left ( \frac{1}{e^x-1}-\frac{1}{e^x+1} \right )d(e^x)\)

\(\Leftrightarrow B=\frac{1}{2}\left.\begin{matrix} 1\\ \frac{1}{2}\end{matrix}\right|\left | \frac{e^x-1}{e^x+1} \right |\approx 0.317\)

c)

Có \(C=\int ^{1}_{0}\frac{(x+2)\ln(x+1)}{(x+1)^2}d(x+1)\).

Đặt \(x+1=t\)

\(\Rightarrow C=\int ^{2}_{1}\frac{(t+1)\ln t}{t^2}dt=\int ^{2}_{1}\frac{\ln t}{t}dt+\int ^{2}_{1}\frac{\ln t}{t^2}dt\)

\(=\int ^{2}_{1}\ln td(\ln t)+\int ^{2}_{1}\frac{\ln t}{t^2}dt=\frac{\ln ^22}{2}+\int ^{2}_{1}\frac{\ln t}{t^2}dt\)

Đặt \(\left\{\begin{matrix} u=\ln t\\ dv=\frac{dt}{t^2}\end{matrix}\right.\Rightarrow \left\{\begin{matrix} du=\frac{dt}{t}\\ v=\frac{-1}{t}\end{matrix}\right.\Rightarrow \int ^{2}_{1}\frac{\ln t}{t^2}dt=\left.\begin{matrix} 2\\ 1\end{matrix}\right|-\frac{\ln t+1}{t}=\frac{1}{2}-\frac{\ln 2 }{2}\)

\(\Rightarrow C=\frac{1}{2}-\frac{\ln 2}{2}+\frac{\ln ^22}{2}\)

d)

\(D=\int ^{\frac{\pi}{4}}_{0}\frac{x\sin x+(x+1)\cos x}{x\sin x+\cos x}dx=\int ^{\frac{\pi}{4}}_{0}dx+\int ^{\frac{\pi}{4}}_{0}\frac{x\cos x}{x\sin x+\cos x}dx\)

Ta có:

\(\int ^{\frac{\pi}{4}}_{0}dx=\left.\begin{matrix} \frac{\pi}{4}\\ 0\end{matrix}\right|x=\frac{\pi}{4}\)

\(\int ^{\frac{\pi}{4}}_{0}\frac{x\cos xdx}{x\sin x+\cos x}=\int ^{\frac{\pi}{4}}_{0}\frac{d(x\sin x+\cos x)}{x\sin x+\cos x}=\left.\begin{matrix} \frac{\pi}{4}\\ 0\end{matrix}\right|\ln |x\sin x+\cos x|\)

\(=\ln|\frac{\pi\sqrt{2}}{8}+\frac{\sqrt{2}}{2}|\)

Suy ra \(D=\frac{\pi}{4}+\ln|\frac{\pi\sqrt{2}}{8}+\frac{\sqrt{2}}{2}|\)

a) Áp dụng phương pháp tìm nguyên hàm từng phần:

Đặt u= ln(1+x)

dv= xdx

=> ,

Ta có: ∫xln(1+x)dx =

=

b) Cách 1: Tìm nguyên hàm từng phần hai lần:

Đặt u= (x²+2x -1) và dv=e^xdx

Suy ra du = (2x+2)dx, v = e^x

. Khi đó:

∫(x²+2x - 1)e^xdx = (x²+2x - 1)e^xdx - ∫(2x+2)e^xdx

Đặt : u=2x+2; dv=e^xdx

=> du = 2dx ;v=e^x

Khi đó:∫(2x+2)e^xdx = (2x+2)e^x - 2∫e^xdx = e^x(2x+2) – 2e^x+C

Vậy

∫(x²+2x+1)e^xdx = e^x(x²-1) + C

Cách 2: HD: Ta tìm ∫(x²-1)e^xdx. Đặt u = x²-1 và dv=e^xdx.

Đáp số : e^x(x²-1) + C

c) Đáp số:

HD: Đặt u=x ; dv = sin(2x+1)dx

d) Đáp số : (1-x)sinx - cosx +C.

HD: Đặt u = 1 - x ;dv = cosxdx

\(=\int\left(6x^2-\dfrac{4}{x}+sin3x-cos4x+e^{2x+1}+9^{x-1}+\dfrac{1}{cos^2x}-\dfrac{1}{sin^2x}\right)dx\)

\(=2x^3-4ln\left|x\right|-\dfrac{1}{3}cos3x-\dfrac{1}{4}sin4x+\dfrac{1}{2}e^{2x+1}+\dfrac{9^{x-1}}{ln9}+tanx+cotx+C\)

Câu 2)

Đặt \(\left\{\begin{matrix} u=\ln ^2x\\ dv=x^2dx\end{matrix}\right.\Rightarrow \left\{\begin{matrix} du=2\frac{\ln x}{x}dx\\ v=\frac{x^3}{3}\end{matrix}\right.\Rightarrow I=\frac{x^3}{3}\ln ^2x-\frac{2}{3}\int x^2\ln xdx\)

Đặt \(\left\{\begin{matrix} k=\ln x\\ dt=x^2dx\end{matrix}\right.\Rightarrow \left\{\begin{matrix} dk=\frac{dx}{x}\\ t=\frac{x^3}{3}\end{matrix}\right.\Rightarrow \int x^2\ln xdx=\frac{x^3\ln x}{3}-\int \frac{x^2}{3}dx=\frac{x^3\ln x}{3}-\frac{x^3}{9}+c\)

Do đó \(I=\frac{x^3\ln^2x}{3}-\frac{2}{9}x^3\ln x+\frac{2}{27}x^3+c\)

Câu 3:

\(I=\int\frac{2}{\cos 2x-7}dx=-\int\frac{2}{2\sin^2x+6}dx=-\int\frac{dx}{\sin^2x+3}\)

Đặt \(t=\tan\frac{x}{2}\Rightarrow \left\{\begin{matrix} \sin x=\frac{2t}{t^2+1}\\ dx=\frac{2dt}{t^2+1}\end{matrix}\right.\)

\(\Rightarrow I=-\int \frac{2dt}{(t^2+1)\left ( \frac{4t^2}{(t^2+1)^2}+3 \right )}=-\int\frac{2(t^2+1)dt}{3t^4+10t^2+3}=-\int \frac{2d\left ( t-\frac{1}{t} \right )}{3\left ( t-\frac{1}{t} \right )^2+16}=\int\frac{2dk}{3k^2+16}\)

Đặt \(k=\frac{4}{\sqrt{3}}\tan v\). Đến đây dễ dàng suy ra \(I=\frac{-1}{2\sqrt{3}}v+c\)

Câu 6)

\(I=-\int \frac{\left ( 1-\frac{1}{x^2} \right )dx}{x^2+2+\frac{1}{x^2}}=-\int \frac{d\left ( x+\frac{1}{x} \right )}{\left ( x+\frac{1}{x} \right )^2}=-\frac{1}{x+\frac{1}{x}}+c=-\frac{x}{x^2+1}+c\)

Câu 8)

\(I=\int \ln \left(\frac{x+1}{x-1}\right)dx=\int \ln (x+1)dx-\int \ln (x-1)dx\)

\(\Leftrightarrow I=\int \ln (x+1)d(x+1)-\int \ln (x-1)d(x-1)\)

Xét \(\int \ln tdt\) ta có:

Đặt \(\left\{\begin{matrix} u=\ln t\\ dv=dt\end{matrix}\right.\Rightarrow \left\{\begin{matrix} du=\frac{dt}{t}\\ v=t\end{matrix}\right.\Rightarrow \int \ln tdt=t\ln t-\int dt=t\ln t-t+c\)

\(\Rightarrow I=(x+1)\ln (x+1)-(x+1)-(x-1)\ln (x-1)+x-1+c\)

\(\Leftrightarrow I=(x+1)\ln(x+1)-(x-1)\ln(x-1)+c\)

a/ Tích phân này làm sao giải được nhỉ?

b/ Đặt \(\sqrt{x}=t\Rightarrow x=t^2\Rightarrow dx=2t.dt\)

\(I=\int\frac{2t^2.dt}{4-t^4}=\int\left(\frac{1}{2-t^2}-\frac{1}{2+t^2}\right)dt=\frac{1}{2\sqrt{2}}ln\left|\frac{\sqrt{2}+t}{\sqrt{2}-t}\right|+\frac{1}{\sqrt{2}}arctan\frac{\sqrt{2}}{t}+C\)

\(=\frac{1}{2\sqrt{2}}ln\left|\frac{\sqrt{2}+\sqrt{x}}{\sqrt{2}-\sqrt{x}}\right|+\frac{1}{\sqrt{2}}arctan\frac{\sqrt{2}}{\sqrt{x}}+C\)

c/ \(I=\int\frac{\sqrt{1+x^2}}{x^2}.xdx\)

Đặt \(\sqrt{1+x^2}=t\Rightarrow x^2=t^2-1\Rightarrow xdx=tdt\)

\(\Rightarrow I=\int\frac{t^2dt}{t^2-1}=\int\left(1+\frac{1}{t^2-1}\right)dt=t+ln\left|\frac{t-1}{t+1}\right|+C=\sqrt{1+x^2}+ln\left|\frac{\sqrt{1+x^2}-1}{\sqrt{1+x^2}+1}\right|+C\)

d/ Con nguyên hàm này cũng không tính được, chắc bạn ghi nhầm đề