Các dạng Nguyên hàm hợp và bài tập thực hành

- \(\int u^{\alpha} d u=\frac{u^{\alpha+1}}{\alpha+1}+C(\alpha \neq-1)\)

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

- \(\int d u=u+C\)

- \(\int \frac{d u}{u}=\ln |u|+C\)

CHÚ Ý: 

Đổi vi phân \(d x=\frac{d u}{u^{\prime}}\)

Tính nguyên hàm \(\int x\left(x^{2}+19\right)^{20} d x\)

giải:

Có : \(I=\int \underline{x}\left(x^{2}+19\right)^{20} d \underline{x}\)

\(\left\{\begin{array}{l}\left(x^{2}+19\right)^{20} \longrightarrow u^{20} \\ u=x^{2}+19 .\end{array}\right.\)

Nhận xét: \(d x=\frac{d\left(x^{2}+19\right)}{2 x}\)

\(\Leftrightarrow 2 x d x=d\left(x^{2}+19\right)\)

\(\Leftrightarrow x d x=\frac{d\left(x^{2}+19\right)}{2}\)

=> \(I=\int\left(x^{2}+19\right)^{20} \cdot(x d x)\)

\(=\int\left(x^{2}+19\right)^{20} \cdot \frac{d\left(x^{2}+19\right)}{2}\)

\(=\frac{1}{2} \int\left(x^{2}+19\right)^{20} \cdot d\left(x^{2}+19\right)\)

\(=\frac{1}{2} \cdot \frac{\left(x^{2}+19\right)^{21}}{21}+C\)

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

giải:

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:

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

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

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

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

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

Tính nguyên hàm \(\int x^{2} \sqrt[3]{1-x^{3}} d x\)

giải:

\(N x:\left\{\begin{array}{l}\sqrt[3]{1-x^{3}} \longrightarrow \sqrt[3]{u} \rightarrow u^{\frac{1}{3}} \\ u=1-x^{3} \rightarrow d u=-3 x^{2} . d x \rightarrow x^{2} d x=\frac{d u}{-3} .\end{array}\right.\)

\(\Rightarrow I=\int \sqrt[3]{1-x^{3}} \cdot x^{2} d x\)

\(=\frac{-1}{3} \cdot \int u^{\frac{1}{3}} d u\)

\(=-\frac{1}{3} \cdot \frac{3}{4} \cdot u^{\frac{4}{3}}+C\)

\(=-\frac{1}{4} \cdot \sqrt[3]{\left(1-x^{3}\right)^{4}}+C\)

Tìm nguyên hàm của \(\int x \sqrt{x^{2}+4} d x\)

giải:

NX: \(\left\{\begin{array}{l}\sqrt{x^{2}+4} \rightarrow \sqrt{u} \\ u=x^{2}+4 \rightarrow d u=2 x d x \rightarrow x d x=\frac{d u}{2} .\end{array}\right.\)

\(I=\int \sqrt{x^{2}+4} \cdot x d x\)

\(=\int \sqrt{u} \cdot \frac{d u}{2}\)

\(=\frac{1}{2} \int u^{\frac{1}{2}} d u\)

= \(\frac{1}{2} \cdot \frac{u^{\frac{1}{2}+1}}{\frac{1}{2}+1}+C\)

\(=\frac{1}{2} \cdot \frac{2}{3} \cdot u^{\frac{3}{2}}+C\)

= \(=\frac{1}{3} \cdot\left(x^{2}+4\right)^{\frac{3}{2}}+C\)

\(=\frac{1}{3} \cdot \sqrt{\left(x^{2}+4\right)^{3}}+C\)

Tình nguyên hàm của \(\int \frac{x^{2} d x}{1-x^{3}}\)

giải: 

NX: \(\quad\left(1-x^{3}\right)^{\prime}=-3 x^{2}\) \(\Rightarrow d\left(1-x^{3}\right)=-\sqrt[3]{x^{2}} \cdot d x\)

\(\Rightarrow I=-\frac{1}{3} \int \frac{d\left(1-x^{3}\right)}{1-x^{3}}=-\frac{1}{3} \cdot \ln \left|1-x^{3}\right|+C\)

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

giải:

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

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

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

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

- \(\int \sin u \cdot d u=-\cos u+C \quad \cdot \int \cos u \cdot d u=\sin u+C\)

- \(\int \frac{d u}{\cos ^{2} u}=\tan u+C \quad \cdot \int \frac{d u}{\sin ^{2} u}=-\cot u+C\)

Tính nguyên hàm \(\int \sin (2 x+1) d x\)

giải:

NX : \(d x=\frac{d(2 x+1)}{2}\)

\(\Rightarrow I=\int \sin (2 x+1) \cdot \frac{d(2 x+1)}{2}\)

\(=\frac{1}{2} \int \sin (2 x+1) \cdot d(2 x+1\)

Tính nguyên hàm \(\int \frac{\sin (\ln x)}{x} d x\)

giải:

\(d(\ln x)=\frac{1}{x} \cdot d x\)

\(\Rightarrow I=\int \sin (\ln x) \cdot d(\ln x)\)

\(=-\cos (\ln x)+C\)