Bài tập các dạng nguyên hàm hợp

Tính \(\int 3 x \left( 6 - 5 x^{2} \right ) dx\)

Cách 1:

 \(I=\int\left(18 x-15 x^{3}\right) d x=18 \cdot \frac{x^{2}}{2}-15 \cdot \frac{x^{4}}{4}+C\)

\(=9 x^{2}-\frac{15}{4} x^{4}+C\)

Cách 2:

\(d x=\frac{d\left(6-5 x^{2}\right)}{-10 x}\) <=> \(x \cdot d x=\frac{d\left(6-5 x^{2}\right)}{-20}\)

\(I=3 \cdot \int\left(6-5 x^{2}\right) \cdot \underline{x d x}=3 \cdot \int\left(6-5 x^{2}\right) \cdot \frac{d\left(6-5 x^{2}\right)}{-10}\)

\(=\frac{-3}{10} \cdot \int \frac{\left(6-5 x^{2}\right) d\left(6-5 x^{2}\right)}{d}=\frac{-3}{10} \cdot \frac{\left(6-5 x^{2}\right)^{2}}{2}+C\)

Tính \(\int x^{2}\left(3 x^{3}+5\right)^{7} d x\)

Nhận xét: \(\left\{\begin{array}{l}\left(3 x^{3}+5\right)^{7} \longrightarrow u^{7} \\ u=3 x^{3}+5 \longrightarrow d u=9 x^{2} \cdot d x\end{array}\right.\)

\(d a=d\left(3 x^{3}+5\right)\)

\(I=\int u^{7} \cdot \frac{d u}{9}=\frac{u^{8}}{72}+C\)

\(=\frac{\left(3 x^{3}+5\right)^{2}}{72}+C\)

Gọi F(x) là một nguyên hàm của \(f(x)=\frac{x^{4} d x}{\sqrt{x^{5}+5}}\) và \(F(-1)=\frac{9}{5}\). Tìm F(x)?

\(F(x)=\int \frac{x^{4}}{\sqrt{x^{5}+5}} \cdot d x \quad u=x^{5}+5 \rightarrow d u=5 x^{4} d x\)

\(=\int \frac{d u}{5 \cdot \sqrt{w}}=\frac{1}{5} \cdot 2 \cdot \int \frac{d u}{2 \sqrt{u}}=\frac{2}{5} \cdot \sqrt{u}+C\)

=> \(F(x)=\frac{2}{5} \sqrt{x^{5}+5}+C\)

\( F(-1)=\frac{9}{5} \Leftrightarrow \frac{2}{5} \sqrt{(-1)^{5}+5}+C=\frac{9}{5}\)

\(\Leftrightarrow \frac{4}{5}+C=\frac{9}{5} \Leftrightarrow C=1\)

Gọi F(x) là nguyên hàm của \(f(x)=\frac{x}{\sqrt{8-x^{2}}}\) thoả \(F(2)=0\). Tìm nghiệm phương trình F(x)=x?

Đáp án:

\(\begin{aligned} F(x)=\int \frac{x}{\sqrt{8-x^{2}}} \cdot d x ; \quad \sqrt{8-x^{2}} \rightarrow u & =8-x^{2} \\ \Rightarrow d u & =-2 x d u\end{aligned}\)

\(=\int \frac{d u}{-2 \sqrt{u}}=-\sqrt{u}+c\)

=> \(F(x)=-\sqrt{8-x^{2}}+C\)

\(\begin{aligned} C_{0}: F(2)=O & \Leftrightarrow-\sqrt{8-2^{2}}+C=0 \\ & \Leftrightarrow C=2 \Rightarrow F(x)=-\sqrt{8-x^{2}}+2\end{aligned}\)

\(F(x)=x \Leftrightarrow-\sqrt{8-x^{2}}+2=x\)

\(\Leftrightarrow \sqrt{8-x^{2}}=2-x\)

\(\Leftrightarrow\left\{\begin{array}{l}2-x \geqslant 0 \\ 8-x^{2}=(2-x)^{2}\end{array}\right.\)

\(\Leftrightarrow\left\{\begin{array}{l}x \leqslant 2 \\ -2 x^{2}+4 x+4=0\end{array}\right.\)

\(\Leftrightarrow\left\{\begin{array}{l}x \leq 2 \\ {\left[\begin{array}{l}x=1+\sqrt{3} \\ x=1-\sqrt{3}\end{array}\right.}\end{array}\right.\)

<=> \(x=1-\sqrt{3}\)

Tính \(\int \frac{2 x-3}{x^{2}-3 x+2} d x\)

Đáp án:

\(x^{2}-3 x+2=(x-1)(x-2) \longrightarrow C_{1}\)

\(\left(x^{2}-3 x+2\right)^{\prime}=2 x-3 \longrightarrow c_{2}\).

\(N x: \quad u=x^{2}-3 x+2 \Rightarrow d u=(2 x-3) d x\)

\(\rightarrow \int \frac{d u}{u}=\ln |u|+C=\ln \left|x^{2}-3 x+2\right|+C\)

Tính \(\int \frac{2 x+1}{x^{2}+x+1} d x\)

Đáp án:

\(N x: d\left(x^{2}+x+1\right)=(2 x+1) d x\)

\(\int \frac{d\left(x^{2}+x+1\right)}{x^{2}+x+1}=\ln (\underbrace{x^{2}+x+1}_{\gt 0})+C\)

Cho hàm số f(x) xđịnh trên R \(\backslash\left\{\frac{1}{2}\right\}\) thỏa mãn \(f^{\prime}(x)=\frac{2}{2 x-1}, f(0)=1, f(1)=2\). Gia trị của biểu thức \(f(-1)+f(3)\) bằng ?

Đáp án:

\(f(x)=\int f^{\prime}(x) d x=\int \frac{2}{2 x-1} \cdot d x=\ln |2 x-1|+C\)

\(\Rightarrow f(x)=\left\{\begin{array}{l}\ln (2 x-1)+C_{1} ? ; x\gt \frac{1}{2} \\ \ln (1-2 x)+C_{2} ? ; x\lt \frac{1}{2} .\end{array}\right.\)

 \(\begin{array}{c}f(0)=1 \Leftrightarrow \ln (1-2 \cdot 0)+C_{2}=1 \\ \Leftrightarrow \ln 1+C_{2}=1\end{array}\) => C2=1

\(\begin{array}{l}f(1)=2 \Leftrightarrow \frac{\ln (2.1-1)}{=0}+C_{1}=2 \\ \Leftrightarrow C_{1}=2 \\\end{array}\)

\(\Rightarrow\left\{\begin{array}{l}f(-1)=\ln 3+1 \\ f(3)=\ln 5+2\end{array}\right.\)

=> \(f(-1)+g(3)=3+\ln 15\).

Tính \(\int e^{x} \cos \left(1+2 e^{x}\right) d x\)

Đáp án:

\(d\left(1+2 e^{x}\right)=2 e^{\frac{x}{x}} \cdot d x\)

<=> \(e^{x} \cdot d x=\frac{d\left(1+2 e^{x}\right)}{2}\)

\(I=\int \cos \left(1+2 e^{x}\right) \cdot d \frac{\left(1+2 e^{x}\right)}{2}\)

\(=\frac{1}{2} \cdot \sin \left(1+2 e^{x}\right) d x\)

Tính \(\int 2 e^{x} \sin \left(e^{x+1}\right) d x\)

Đáp án:

\(d\left(e^{x+1}\right)=e^{x+1} \cdot d x\)

\(=e \cdot e^{x} \cdot d x \Leftrightarrow e^{x} d x=\frac{d\left(e^{x+1}\right)}{e}\)

\(I=2 \cdot \frac{1}{e} \int \frac{\sin \left(e^{x+1}\right)}{\sin u} \frac{d\left(e^{x+1}\right)}{d u}\)

\(=\frac{-2}{e} \cdot \cos \left(e^{x+1}\right)+C\)